ISOSCELES TRIANGLES [Including Inequalities] - Questions & Answers
EXERCISE 10(A)1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) In the given figure, ∠B = ∠C and ∠BAD = ∠CAD, then :
(i) AB = AC
(ii) AB ≠ AC
(iii) ∠ADB ≠ ∠ADC
(iv) ∠ADB ≠ 90°
Step 1: Consider ΔABD and ΔACD.
Step 2: We are given that ∠B = ∠C.
Step 3: We are given that ∠BAD = ∠CAD.
Step 4: The side AD is common to both triangles (AD = AD).
Step 5: By Angle-Angle-Side (AAS) congruence criterion, ΔABD ≅ ΔACD.
Step 6: Since the triangles are congruent, their corresponding parts are equal.
Step 7: Therefore, AB = AC.
Answer: (i) AB = AC
Step 2: We are given that ∠B = ∠C.
Step 3: We are given that ∠BAD = ∠CAD.
Step 4: The side AD is common to both triangles (AD = AD).
Step 5: By Angle-Angle-Side (AAS) congruence criterion, ΔABD ≅ ΔACD.
Step 6: Since the triangles are congruent, their corresponding parts are equal.
Step 7: Therefore, AB = AC.
Answer: (i) AB = AC
(b) In the given figure, AD ⊥ BC and AB = AC, then :
(i) ΔABD ≅ ΔACD
(ii) BD = CD
(iii) ∠BAC = 90°
(iv) ∠CAD = 45°
Step 1: Consider the right-angled triangles ΔABD and ΔACD.
Step 2: The hypotenuses are equal because AB = AC (Given).
Step 3: The side AD is common to both triangles (AD = AD).
Step 4: Since AD ⊥ BC, ∠ADB = ∠ADC = 90°.
Step 5: By Right-Angle Hypotenuse Side (RHS) congruence criterion, ΔABD ≅ ΔACD.
Answer: (i) ΔABD ≅ ΔACD
Step 2: The hypotenuses are equal because AB = AC (Given).
Step 3: The side AD is common to both triangles (AD = AD).
Step 4: Since AD ⊥ BC, ∠ADB = ∠ADC = 90°.
Step 5: By Right-Angle Hypotenuse Side (RHS) congruence criterion, ΔABD ≅ ΔACD.
Answer: (i) ΔABD ≅ ΔACD
(c) In the given figure, AD = BD, then angle ACD is :
(i) 43°
(ii) 22°
(iii) 65°
(iv) 28°
Step 1: In ΔABD, it is given that AD = BD.
Step 2: Angles opposite to equal sides are equal, so ∠BAD = ∠ABD = 65°.
Step 3: The exterior angle of ΔABD at D is ∠ADC.
Step 4: An exterior angle is equal to the sum of opposite interior angles, so ∠ADC = ∠BAD + ∠ABD = 65° + 65° = 130°.
Step 5: In ΔACD, the sum of all internal angles is 180°.
Step 6: ∠ACD + ∠CAD + ∠ADC = 180°.
Step 7: ∠ACD + 22° + 130° = 180°.
Step 8: ∠ACD = 180° - 152° = 28°.
Answer: (iv) 28°
Step 2: Angles opposite to equal sides are equal, so ∠BAD = ∠ABD = 65°.
Step 3: The exterior angle of ΔABD at D is ∠ADC.
Step 4: An exterior angle is equal to the sum of opposite interior angles, so ∠ADC = ∠BAD + ∠ABD = 65° + 65° = 130°.
Step 5: In ΔACD, the sum of all internal angles is 180°.
Step 6: ∠ACD + ∠CAD + ∠ADC = 180°.
Step 7: ∠ACD + 22° + 130° = 180°.
Step 8: ∠ACD = 180° - 152° = 28°.
Answer: (iv) 28°
(d) In the given figure; BE = DC, then :
(i) AD = DC
(ii) AE = BE
(iii) AD = AE
(iv) ∠ABE = ∠DAC
Step 1: In ΔABC, the markings indicate that side AB = side AC.
Step 2: Therefore, the angles opposite to these sides are equal, meaning ∠B = ∠C.
Step 3: Now, consider ΔABE and ΔACD.
Step 4: AB = AC (Given from markings).
Step 5: ∠B = ∠C (Proved in Step 2).
Step 6: BE = DC (Given).
Step 7: By Side-Angle-Side (SAS) congruence criterion, ΔABE ≅ ΔACD.
Step 8: By Corresponding Parts of Congruent Triangles are Congruent (CPCTC), AD = AE.
Answer: (iii) AD = AE
Step 2: Therefore, the angles opposite to these sides are equal, meaning ∠B = ∠C.
Step 3: Now, consider ΔABE and ΔACD.
Step 4: AB = AC (Given from markings).
Step 5: ∠B = ∠C (Proved in Step 2).
Step 6: BE = DC (Given).
Step 7: By Side-Angle-Side (SAS) congruence criterion, ΔABE ≅ ΔACD.
Step 8: By Corresponding Parts of Congruent Triangles are Congruent (CPCTC), AD = AE.
Answer: (iii) AD = AE
(e) In ΔABC and ΔPQR, AB = AC, ∠C = ∠P and ∠B = ∠Q; then triangles are :
(i) isosceles but not congruent
(ii) isosceles and congruent
(iii) congruent but not isosceles
(iv) neither isosceles nor congruent.
Step 1: In ΔABC, it is given that AB = AC.
Step 2: Angles opposite to equal sides are equal, so ∠B = ∠C.
Step 3: It is given that ∠C = ∠P and ∠B = ∠Q.
Step 4: Since ∠B = ∠C, it logically follows that ∠P = ∠Q.
Step 5: In ΔPQR, since ∠P = ∠Q, the sides opposite to them are equal (QR = PR).
Step 6: This makes ΔPQR an isosceles triangle.
Step 7: However, no corresponding side lengths between the two triangles are provided, so we cannot prove they are congruent.
Answer: (i) isosceles but not congruent
Step 2: Angles opposite to equal sides are equal, so ∠B = ∠C.
Step 3: It is given that ∠C = ∠P and ∠B = ∠Q.
Step 4: Since ∠B = ∠C, it logically follows that ∠P = ∠Q.
Step 5: In ΔPQR, since ∠P = ∠Q, the sides opposite to them are equal (QR = PR).
Step 6: This makes ΔPQR an isosceles triangle.
Step 7: However, no corresponding side lengths between the two triangles are provided, so we cannot prove they are congruent.
Answer: (i) isosceles but not congruent
2. In the figure alongside, AB = AC, ∠A = 48° and ∠ACD = 18°. Show that : BC = CD.
Step 1: In ΔABC, we are given that AB = AC.
Step 2: Therefore, the angles opposite to these sides are equal: ∠B = ∠ACB.
Step 3: The sum of angles in ΔABC is 180°, so ∠B + ∠ACB + ∠A = 180°.
Step 4: 2∠B + 48° = 180°.
Step 5: 2∠B = 132° ⇒ ∠B = 66°. Thus, ∠ACB = 66°.
Step 6: From the given figure, ∠BCD = ∠ACB - ∠ACD.
Step 7: ∠BCD = 66° - 18° = 48°.
Step 8: In ΔBCD, the sum of angles is 180°, so ∠BDC = 180° - (∠B + ∠BCD).
Step 9: ∠BDC = 180° - (66° + 48°) = 180° - 114° = 66°.
Step 10: Since ∠B = ∠BDC = 66°, the sides opposite to them in ΔBCD must be equal.
Step 11: Therefore, BC = CD. Hence Proved.
Step 2: Therefore, the angles opposite to these sides are equal: ∠B = ∠ACB.
Step 3: The sum of angles in ΔABC is 180°, so ∠B + ∠ACB + ∠A = 180°.
Step 4: 2∠B + 48° = 180°.
Step 5: 2∠B = 132° ⇒ ∠B = 66°. Thus, ∠ACB = 66°.
Step 6: From the given figure, ∠BCD = ∠ACB - ∠ACD.
Step 7: ∠BCD = 66° - 18° = 48°.
Step 8: In ΔBCD, the sum of angles is 180°, so ∠BDC = 180° - (∠B + ∠BCD).
Step 9: ∠BDC = 180° - (66° + 48°) = 180° - 114° = 66°.
Step 10: Since ∠B = ∠BDC = 66°, the sides opposite to them in ΔBCD must be equal.
Step 11: Therefore, BC = CD. Hence Proved.
3. Calculate : (i) ∠ADC (ii) ∠ABC (iii) ∠BAC
Step 1: In the given figure, the exterior angle at C is 130°.
Step 2: The interior adjacent angle is ∠ACD = 180° - 130° = 50° (Linear pair).
Step 3: In ΔADC, the markings indicate AC = DC.
Step 4: Therefore, angles opposite to equal sides are equal: ∠CAD = ∠ADC.
Step 5: Sum of angles in ΔADC = 180° ⇒ 2∠ADC + 50° = 180° ⇒ 2∠ADC = 130° ⇒ ∠ADC = 65°.
Answer (i): ∠ADC = 65°
Step 6: In ΔABC, the markings indicate AB = AC.
Step 7: Therefore, angles opposite to equal sides are equal: ∠ABC = ∠ACB.
Step 8: From the figure, ∠ACB is the same angle as ∠ACD, which is 50°.
Step 9: Thus, ∠ABC = 50°.
Answer (ii): ∠ABC = 50°
Step 10: In ΔABC, sum of angles = 180° ⇒ ∠BAC + 50° + 50° = 180°.
Step 11: ∠BAC = 180° - 100° = 80°.
Answer (iii): ∠BAC = 80°
Step 2: The interior adjacent angle is ∠ACD = 180° - 130° = 50° (Linear pair).
Step 3: In ΔADC, the markings indicate AC = DC.
Step 4: Therefore, angles opposite to equal sides are equal: ∠CAD = ∠ADC.
Step 5: Sum of angles in ΔADC = 180° ⇒ 2∠ADC + 50° = 180° ⇒ 2∠ADC = 130° ⇒ ∠ADC = 65°.
Answer (i): ∠ADC = 65°
Step 6: In ΔABC, the markings indicate AB = AC.
Step 7: Therefore, angles opposite to equal sides are equal: ∠ABC = ∠ACB.
Step 8: From the figure, ∠ACB is the same angle as ∠ACD, which is 50°.
Step 9: Thus, ∠ABC = 50°.
Answer (ii): ∠ABC = 50°
Step 10: In ΔABC, sum of angles = 180° ⇒ ∠BAC + 50° + 50° = 180°.
Step 11: ∠BAC = 180° - 100° = 80°.
Answer (iii): ∠BAC = 80°
5. Calculate x :
(i)
Step 1: In the given triangle, the markings show that two sides are equal.
Step 2: The angle opposite to one of these equal sides is given as 37°.
Step 3: Therefore, the angle opposite to the other equal side is also 37°.
Step 4: The angle 'x' is an exterior angle to this triangle.
Step 5: The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Step 6: x = 37° + 37°.
Step 7: x = 74°.
Answer: x = 74°
Step 2: The angle opposite to one of these equal sides is given as 37°.
Step 3: Therefore, the angle opposite to the other equal side is also 37°.
Step 4: The angle 'x' is an exterior angle to this triangle.
Step 5: The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
Step 6: x = 37° + 37°.
Step 7: x = 74°.
Answer: x = 74°
(ii)
Step 1: In the given triangle, the markings show that two sides (left and right) are equal, so it is an isosceles triangle.
Step 2: The base angles opposite these sides must be equal.
Step 3: The angle on the left is given as 50°, so the angle on the right (interior) is also 50°.
Step 4: The angle 'x' is the exterior angle adjacent to this 50° interior angle.
Step 5: The sum of an interior angle and its adjacent exterior angle on a straight line is 180°.
Step 6: x + 50° = 180°.
Step 7: x = 180° - 50° = 130°.
Answer: x = 130°
Step 2: The base angles opposite these sides must be equal.
Step 3: The angle on the left is given as 50°, so the angle on the right (interior) is also 50°.
Step 4: The angle 'x' is the exterior angle adjacent to this 50° interior angle.
Step 5: The sum of an interior angle and its adjacent exterior angle on a straight line is 180°.
Step 6: x + 50° = 180°.
Step 7: x = 180° - 50° = 130°.
Answer: x = 130°
6. In the figure, given below, AB = AC. Prove that : ∠BOC = ∠ACD.
Step 1: In ΔABC, it is given that AB = AC.
Step 2: Therefore, the angles opposite these sides are equal: ∠ABC = ∠ACB.
Step 3: The markings show that OB and OC are the angle bisectors of ∠ABC and ∠ACB respectively.
Step 4: Thus, ∠OBC = ½∠ABC and ∠OCB = ½∠ACB.
Step 5: Since ∠ABC = ∠ACB, it follows that ∠OBC = ∠OCB.
Step 6: In ΔOBC, the sum of angles is 180°, so ∠BOC = 180° - (∠OBC + ∠OCB) = 180° - 2(½∠ACB) = 180° - ∠ACB.
Step 7: From the figure, ∠ACD is the exterior angle at vertex C, which lies on the straight line B-C-D.
Step 8: Therefore, ∠ACD + ∠ACB = 180° (Linear pair).
Step 9: This means ∠ACD = 180° - ∠ACB.
Step 10: Comparing Step 6 and Step 9, we get ∠BOC = ∠ACD. Hence Proved.
Step 2: Therefore, the angles opposite these sides are equal: ∠ABC = ∠ACB.
Step 3: The markings show that OB and OC are the angle bisectors of ∠ABC and ∠ACB respectively.
Step 4: Thus, ∠OBC = ½∠ABC and ∠OCB = ½∠ACB.
Step 5: Since ∠ABC = ∠ACB, it follows that ∠OBC = ∠OCB.
Step 6: In ΔOBC, the sum of angles is 180°, so ∠BOC = 180° - (∠OBC + ∠OCB) = 180° - 2(½∠ACB) = 180° - ∠ACB.
Step 7: From the figure, ∠ACD is the exterior angle at vertex C, which lies on the straight line B-C-D.
Step 8: Therefore, ∠ACD + ∠ACB = 180° (Linear pair).
Step 9: This means ∠ACD = 180° - ∠ACB.
Step 10: Comparing Step 6 and Step 9, we get ∠BOC = ∠ACD. Hence Proved.
7. In the figure given below, LM = LN; angle PLN = 110°. Calculate : (i) ∠LMN (ii) ∠MLN
Step 1: Assuming points P, L, and M form a straight line, the exterior angle ∠PLN = 110°.
Step 2: The interior adjacent angle ∠MLN and exterior angle ∠PLN form a linear pair.
Step 3: ∠MLN + ∠PLN = 180°.
Step 4: ∠MLN = 180° - 110° = 70°.
Answer (ii): ∠MLN = 70°
Step 5: In ΔLMN, we are given that LM = LN.
Step 6: Therefore, the angles opposite these sides are equal: ∠LMN = ∠LNM.
Step 7: The sum of angles in ΔLMN is 180°.
Step 8: ∠LMN + ∠LNM + ∠MLN = 180°.
Step 9: 2∠LMN + 70° = 180°.
Step 10: 2∠LMN = 110° ⇒ ∠LMN = 55°.
Answer (i): ∠LMN = 55°
Step 2: The interior adjacent angle ∠MLN and exterior angle ∠PLN form a linear pair.
Step 3: ∠MLN + ∠PLN = 180°.
Step 4: ∠MLN = 180° - 110° = 70°.
Answer (ii): ∠MLN = 70°
Step 5: In ΔLMN, we are given that LM = LN.
Step 6: Therefore, the angles opposite these sides are equal: ∠LMN = ∠LNM.
Step 7: The sum of angles in ΔLMN is 180°.
Step 8: ∠LMN + ∠LNM + ∠MLN = 180°.
Step 9: 2∠LMN + 70° = 180°.
Step 10: 2∠LMN = 110° ⇒ ∠LMN = 55°.
Answer (i): ∠LMN = 55°
8. An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°. Find : (i) ∠DCB (ii) ∠CBD.
Step 1: In ΔABC, we are given AC = BC, which means ∠CAB = ∠CBA.
Step 2: We are given ∠CAB = 55°, therefore ∠CBA (which is the same as ∠CBD) = 55°.
Answer (ii): ∠CBD = 55°
Step 3: In an isosceles triangle, the median drawn to the unequal base acts as an altitude as well.
Step 4: Since CD bisects AB at D, CD is the median, so CD ⊥ AB.
Step 5: Therefore, ∠CDB = 90°.
Step 6: In ΔCDB, the sum of angles is 180°.
Step 7: ∠DCB + ∠CBD + ∠CDB = 180°.
Step 8: ∠DCB + 55° + 90° = 180°.
Step 9: ∠DCB = 180° - 145° = 35°.
Answer (i): ∠DCB = 35°
Step 2: We are given ∠CAB = 55°, therefore ∠CBA (which is the same as ∠CBD) = 55°.
Answer (ii): ∠CBD = 55°
Step 3: In an isosceles triangle, the median drawn to the unequal base acts as an altitude as well.
Step 4: Since CD bisects AB at D, CD is the median, so CD ⊥ AB.
Step 5: Therefore, ∠CDB = 90°.
Step 6: In ΔCDB, the sum of angles is 180°.
Step 7: ∠DCB + ∠CBD + ∠CDB = 180°.
Step 8: ∠DCB + 55° + 90° = 180°.
Step 9: ∠DCB = 180° - 145° = 35°.
Answer (i): ∠DCB = 35°
9. Find x :
Step 1: In the given figure, there is a triangle with an exterior angle marked as 'x'.
Step 2: The interior angle at the opposite vertex is marked as 42°.
Step 3: The markings indicate that the side opposite the 42° angle is equal to the base of the top sub-triangle.
Step 4: Since the triangle is isosceles, the base angles are equal. So the interior angle adjacent to the base is 42°.
Step 5: The third angle inside the main triangle is 180° - (42° + 42°) = 180° - 84° = 96°.
Step 6: The angle 'x' forms a straight line with the interior angle of 42°, or it serves as the exterior angle.
Step 7: Based on the geometric setup, x is the exterior angle formed by extending the side.
Step 8: The exterior angle is the sum of opposite interior angles: x = 42° + 96° = 138°.
Answer: x = 138°
Step 2: The interior angle at the opposite vertex is marked as 42°.
Step 3: The markings indicate that the side opposite the 42° angle is equal to the base of the top sub-triangle.
Step 4: Since the triangle is isosceles, the base angles are equal. So the interior angle adjacent to the base is 42°.
Step 5: The third angle inside the main triangle is 180° - (42° + 42°) = 180° - 84° = 96°.
Step 6: The angle 'x' forms a straight line with the interior angle of 42°, or it serves as the exterior angle.
Step 7: Based on the geometric setup, x is the exterior angle formed by extending the side.
Step 8: The exterior angle is the sum of opposite interior angles: x = 42° + 96° = 138°.
Answer: x = 138°
10. In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.
Step 1: Consider ΔABD and ΔCBD.
Step 2: Since BD bisects ∠B, ∠ABD = ∠CBD.
Step 3: Since BD ⊥ AC, ∠BDA = ∠BDC = 90°.
Step 4: The side BD is common to both triangles (BD = BD).
Step 5: By Angle-Side-Angle (ASA) congruence criterion, ΔABD ≅ ΔCBD.
Step 6: By CPCTC, corresponding sides are equal: AB = BC and AD = CD.
Step 7: From the given figure, AB = 3x + 1 and BC = 5y - 2, so 3x + 1 = 5y - 2.
Step 8: Rearranging this equation gives: 3x - 5y = -3 (Equation 1).
Step 9: From the given figure, AD = x + 1 and CD = y + 2, so x + 1 = y + 2.
Step 10: Rearranging this equation gives: x - y = 1 ⇒ x = y + 1 (Equation 2).
Step 11: Substitute Equation 2 into Equation 1: 3(y + 1) - 5y = -3.
Step 12: 3y + 3 - 5y = -3 ⇒ -2y = -6 ⇒ y = 3.
Step 13: Substitute y = 3 back into Equation 2: x = 3 + 1 = 4.
Answer: x = 4, y = 3
Step 2: Since BD bisects ∠B, ∠ABD = ∠CBD.
Step 3: Since BD ⊥ AC, ∠BDA = ∠BDC = 90°.
Step 4: The side BD is common to both triangles (BD = BD).
Step 5: By Angle-Side-Angle (ASA) congruence criterion, ΔABD ≅ ΔCBD.
Step 6: By CPCTC, corresponding sides are equal: AB = BC and AD = CD.
Step 7: From the given figure, AB = 3x + 1 and BC = 5y - 2, so 3x + 1 = 5y - 2.
Step 8: Rearranging this equation gives: 3x - 5y = -3 (Equation 1).
Step 9: From the given figure, AD = x + 1 and CD = y + 2, so x + 1 = y + 2.
Step 10: Rearranging this equation gives: x - y = 1 ⇒ x = y + 1 (Equation 2).
Step 11: Substitute Equation 2 into Equation 1: 3(y + 1) - 5y = -3.
Step 12: 3y + 3 - 5y = -3 ⇒ -2y = -6 ⇒ y = 3.
Step 13: Substitute y = 3 back into Equation 2: x = 3 + 1 = 4.
Answer: x = 4, y = 3
11. In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.
Step 1: Let the angles be ∠A = 8x and ∠B = 5x.
Step 2: In ΔABC, we are given that AB = AC.
Step 3: Angles opposite to equal sides are equal, so ∠C = ∠B = 5x.
Step 4: The sum of angles in a triangle is 180°.
Step 5: ∠A + ∠B + ∠C = 180°.
Step 6: 8x + 5x + 5x = 180°.
Step 7: 18x = 180° ⇒ x = 10°.
Step 8: Therefore, ∠A = 8x = 8 × 10° = 80°.
Answer: ∠A = 80°
Step 2: In ΔABC, we are given that AB = AC.
Step 3: Angles opposite to equal sides are equal, so ∠C = ∠B = 5x.
Step 4: The sum of angles in a triangle is 180°.
Step 5: ∠A + ∠B + ∠C = 180°.
Step 6: 8x + 5x + 5x = 180°.
Step 7: 18x = 180° ⇒ x = 10°.
Step 8: Therefore, ∠A = 8x = 8 × 10° = 80°.
Answer: ∠A = 80°
12. In triangle ABC; ∠A = 60°, ∠C = 40° and bisector of angle ABC meets side AC at point P. Show that BP = CP.
Step 1: The sum of angles in ΔABC is 180°.
Step 2: ∠A + ∠ABC + ∠C = 180°.
Step 3: 60° + ∠ABC + 40° = 180°.
Step 4: ∠ABC = 180° - 100° = 80°.
Step 5: BP is the bisector of ∠ABC, so ∠PBC = ∠ABC / 2.
Step 6: ∠PBC = 80° / 2 = 40°.
Step 7: Now, in ΔBPC, ∠PBC = 40° and ∠C = 40°.
Step 8: Since ∠PBC = ∠C, the sides opposite to these angles must be equal.
Step 9: Therefore, BP = CP. Hence Proved.
Step 2: ∠A + ∠ABC + ∠C = 180°.
Step 3: 60° + ∠ABC + 40° = 180°.
Step 4: ∠ABC = 180° - 100° = 80°.
Step 5: BP is the bisector of ∠ABC, so ∠PBC = ∠ABC / 2.
Step 6: ∠PBC = 80° / 2 = 40°.
Step 7: Now, in ΔBPC, ∠PBC = 40° and ∠C = 40°.
Step 8: Since ∠PBC = ∠C, the sides opposite to these angles must be equal.
Step 9: Therefore, BP = CP. Hence Proved.
EXERCISE 10(B)
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) In the given figure, AB ⊥ BE, EF ⊥ BE, AB = EF and BC = DE, then :
(i) ΔABD ≅ ΔEFC
(ii) ΔABD ≅ ΔFEC
(iii) ΔABD ≅ ΔECF
(iv) ΔABD ≅ ΔCEF
Step 1: Consider ΔABD and ΔFEC.
Step 2: It is given that AB ⊥ BE and EF ⊥ BE, so ∠B = ∠E = 90°.
Step 3: It is given that AB = EF.
Step 4: We are given BC = DE. By adding CD to both sides, we get BC + CD = DE + CD.
Step 5: This gives BD = CE.
Step 6: By Side-Angle-Side (SAS) congruence, ΔABD ≅ ΔFEC.
Answer: (ii) ΔABD ≅ ΔFEC
Step 2: It is given that AB ⊥ BE and EF ⊥ BE, so ∠B = ∠E = 90°.
Step 3: It is given that AB = EF.
Step 4: We are given BC = DE. By adding CD to both sides, we get BC + CD = DE + CD.
Step 5: This gives BD = CE.
Step 6: By Side-Angle-Side (SAS) congruence, ΔABD ≅ ΔFEC.
Answer: (ii) ΔABD ≅ ΔFEC
(b) From the adjoining figure, we find :
(i) OP = OR
(ii) OP = OQ
(iii) PQ = PR
(iv) PR ≠ PQ
Step 1: The figure shows a line segment intersecting a coordinate-like axis.
Step 2: Right angles are marked at the axis, and markings show distance equality from the center.
Step 3: Because the perpendicular distance from the center point to the ends is the same, this creates symmetric right-angled triangles.
Step 4: Therefore, the hypotenuses generated are equal, making OP = OQ.
Answer: (ii) OP = OQ
Step 2: Right angles are marked at the axis, and markings show distance equality from the center.
Step 3: Because the perpendicular distance from the center point to the ends is the same, this creates symmetric right-angled triangles.
Step 4: Therefore, the hypotenuses generated are equal, making OP = OQ.
Answer: (ii) OP = OQ
(c) From the given figure, if ∠A = ∠C, we get :
(i) x = 8, y = 16
(ii) x = -8, y = 16
(iii) x = 16, y = -8
(iv) x = 16, y = 8
Step 1: In ΔABC, we are given ∠A = ∠C.
Step 2: Therefore, the sides opposite these angles are equal, meaning BC = AB.
Step 3: From the figure, the sides are marked as expressions of x and y, and an altitude is drawn.
Step 4: Base parts are x and 2y. The sides are 2x and 3y + 8.
Step 5: Using the property of an altitude in an isosceles triangle, it bisects the base, so x = 2y.
Step 6: Therefore, x - 2y = 0.
Step 7: Additionally, the equal sides give 2x = 3y + 8.
Step 8: Substitute x = 2y into the sides equation: 2(2y) = 3y + 8.
Step 9: 4y = 3y + 8 ⇒ y = 8.
Step 10: Substitute y = 8 back into x = 2y: x = 2(8) = 16.
Answer: (iv) x = 16, y = 8
Step 2: Therefore, the sides opposite these angles are equal, meaning BC = AB.
Step 3: From the figure, the sides are marked as expressions of x and y, and an altitude is drawn.
Step 4: Base parts are x and 2y. The sides are 2x and 3y + 8.
Step 5: Using the property of an altitude in an isosceles triangle, it bisects the base, so x = 2y.
Step 6: Therefore, x - 2y = 0.
Step 7: Additionally, the equal sides give 2x = 3y + 8.
Step 8: Substitute x = 2y into the sides equation: 2(2y) = 3y + 8.
Step 9: 4y = 3y + 8 ⇒ y = 8.
Step 10: Substitute y = 8 back into x = 2y: x = 2(8) = 16.
Answer: (iv) x = 16, y = 8
(d) ABCD is a rectangle. X and Y are points on sides AD and BC respectively such that AX = BY, then :
(i) AY ≠ BX
(ii) ΔABX ≅ ΔBYA
(iii) ΔABX ≅ ΔAYB
(iv) ΔABX ≅ ΔBAY
Step 1: Consider ΔABX and ΔBAY.
Step 2: Since ABCD is a rectangle, ∠A = ∠B = 90°.
Step 3: It is given that AX = BY.
Step 4: The base AB is common to both triangles (AB = BA).
Step 5: By Side-Angle-Side (SAS) congruence criterion, ΔABX ≅ ΔBAY.
Answer: (iv) ΔABX ≅ ΔBAY
Step 2: Since ABCD is a rectangle, ∠A = ∠B = 90°.
Step 3: It is given that AX = BY.
Step 4: The base AB is common to both triangles (AB = BA).
Step 5: By Side-Angle-Side (SAS) congruence criterion, ΔABX ≅ ΔBAY.
Answer: (iv) ΔABX ≅ ΔBAY
(e) In the given figure, P is mid-point of side AD of rectangle ABCD; then :
(i) ∠PBC = ∠PBA
(ii) ∠PBC = ∠PCB
(iii) ∠BPA = ∠BPC
(iv) ∠PBC = ∠BPA
Step 1: Consider ΔPAB and ΔPDC.
Step 2: Since ABCD is a rectangle, ∠A = ∠D = 90° and AB = DC.
Step 3: P is the midpoint of AD, so PA = PD.
Step 4: By SAS congruence criterion, ΔPAB ≅ ΔPDC.
Step 5: By CPCTC, PB = PC.
Step 6: In ΔPBC, since PB = PC, the angles opposite these sides are equal.
Step 7: Therefore, ∠PBC = ∠PCB.
Answer: (ii) ∠PBC = ∠PCB
Step 2: Since ABCD is a rectangle, ∠A = ∠D = 90° and AB = DC.
Step 3: P is the midpoint of AD, so PA = PD.
Step 4: By SAS congruence criterion, ΔPAB ≅ ΔPDC.
Step 5: By CPCTC, PB = PC.
Step 6: In ΔPBC, since PB = PC, the angles opposite these sides are equal.
Step 7: Therefore, ∠PBC = ∠PCB.
Answer: (ii) ∠PBC = ∠PCB
2. In the given figure, AB = AC. Prove that : (i) DP = DQ (ii) AP = AQ (iii) AD bisects angle A
Step 1: In ΔABC, AB = AC ⇒ ∠B = ∠C.
Step 2: Consider right-angled triangles ΔDPB and ΔDQC.
Step 3: ∠DPB = ∠DQC = 90°.
Step 4: ∠B = ∠C (Proved).
Step 5: Since D is a point on BC, if we assume BD = CD (midpoint), then by AAS ΔDPB ≅ ΔDQC. (Assuming D is midpoint based on standard figure setup).
Step 6: By CPCTC, DP = DQ. Hence Proved (i).
Step 7: Consider ΔAPD and ΔAQD.
Step 8: ∠APD = ∠AQD = 90°.
Step 9: AD = AD (Common hypotenuse).
Step 10: DP = DQ (Proved in i).
Step 11: By RHS congruence criterion, ΔAPD ≅ ΔAQD.
Step 12: By CPCTC, AP = AQ. Hence Proved (ii).
Step 13: Also by CPCTC, ∠PAD = ∠QAD.
Step 14: This means AD bisects angle A. Hence Proved (iii).
Step 2: Consider right-angled triangles ΔDPB and ΔDQC.
Step 3: ∠DPB = ∠DQC = 90°.
Step 4: ∠B = ∠C (Proved).
Step 5: Since D is a point on BC, if we assume BD = CD (midpoint), then by AAS ΔDPB ≅ ΔDQC. (Assuming D is midpoint based on standard figure setup).
Step 6: By CPCTC, DP = DQ. Hence Proved (i).
Step 7: Consider ΔAPD and ΔAQD.
Step 8: ∠APD = ∠AQD = 90°.
Step 9: AD = AD (Common hypotenuse).
Step 10: DP = DQ (Proved in i).
Step 11: By RHS congruence criterion, ΔAPD ≅ ΔAQD.
Step 12: By CPCTC, AP = AQ. Hence Proved (ii).
Step 13: Also by CPCTC, ∠PAD = ∠QAD.
Step 14: This means AD bisects angle A. Hence Proved (iii).
3. In triangle ABC, AB = AC; BE ⊥ AC and CF ⊥ AB. Prove that : (i) BE = CF (ii) AF = AE
Step 1: Consider ΔABE and ΔACF.
Step 2: ∠A = ∠A (Common angle).
Step 3: ∠AEB = ∠AFC = 90° (Since BE ⊥ AC and CF ⊥ AB).
Step 4: AB = AC (Given).
Step 5: By Angle-Angle-Side (AAS) congruence criterion, ΔABE ≅ ΔACF.
Step 6: By CPCTC, BE = CF. Hence Proved (i).
Step 7: Also by CPCTC, AE = AF, which is the same as AF = AE. Hence Proved (ii).
Step 2: ∠A = ∠A (Common angle).
Step 3: ∠AEB = ∠AFC = 90° (Since BE ⊥ AC and CF ⊥ AB).
Step 4: AB = AC (Given).
Step 5: By Angle-Angle-Side (AAS) congruence criterion, ΔABE ≅ ΔACF.
Step 6: By CPCTC, BE = CF. Hence Proved (i).
Step 7: Also by CPCTC, AE = AF, which is the same as AF = AE. Hence Proved (ii).
6. Prove that the bisectors of the base angles of an isosceles triangle are equal.
Step 1: Let ABC be an isosceles triangle with AB = AC.
Step 2: The base angles are ∠B and ∠C. Since AB = AC, ∠B = ∠C.
Step 3: Let BE and CF be the angle bisectors of ∠B and ∠C respectively.
Step 4: This means ∠EBC = ½∠B and ∠FCB = ½∠C.
Step 5: Since ∠B = ∠C, we have ∠EBC = ∠FCB.
Step 6: Consider ΔBCE and ΔCBF.
Step 7: ∠C = ∠B (Base angles).
Step 8: BC = CB (Common side).
Step 9: ∠EBC = ∠FCB (Proved in Step 5).
Step 10: By Angle-Side-Angle (ASA) congruence criterion, ΔBCE ≅ ΔCBF.
Step 11: By CPCTC, BE = CF.
Step 12: Therefore, the bisectors of the base angles are equal. Hence Proved.
Step 2: The base angles are ∠B and ∠C. Since AB = AC, ∠B = ∠C.
Step 3: Let BE and CF be the angle bisectors of ∠B and ∠C respectively.
Step 4: This means ∠EBC = ½∠B and ∠FCB = ½∠C.
Step 5: Since ∠B = ∠C, we have ∠EBC = ∠FCB.
Step 6: Consider ΔBCE and ΔCBF.
Step 7: ∠C = ∠B (Base angles).
Step 8: BC = CB (Common side).
Step 9: ∠EBC = ∠FCB (Proved in Step 5).
Step 10: By Angle-Side-Angle (ASA) congruence criterion, ΔBCE ≅ ΔCBF.
Step 11: By CPCTC, BE = CF.
Step 12: Therefore, the bisectors of the base angles are equal. Hence Proved.
EXERCISE 10(C)
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) In the adjoining figure, we find :
(i) AB = AC
(ii) BC > AB
(iii) AB > BC
(iv) AC = BC
Step 1: In the given triangle, the exterior angle at the top is 115°.
Step 2: The interior adjacent angle is ∠A = 180° - 115° = 65°.
Step 3: The exterior angle at B is 125°, so interior ∠B = 180° - 125° = 55°.
Step 4: Sum of angles = 180°, so ∠C = 180° - (65° + 55°) = 180° - 120° = 60°.
Step 5: ∠A is the largest angle (65°), so the side opposite it, BC, is the largest side.
Step 6: Comparing BC and AB, since ∠A (65°) > ∠C (60°), side BC > AB.
Answer: (ii) BC > AB
Step 2: The interior adjacent angle is ∠A = 180° - 115° = 65°.
Step 3: The exterior angle at B is 125°, so interior ∠B = 180° - 125° = 55°.
Step 4: Sum of angles = 180°, so ∠C = 180° - (65° + 55°) = 180° - 120° = 60°.
Step 5: ∠A is the largest angle (65°), so the side opposite it, BC, is the largest side.
Step 6: Comparing BC and AB, since ∠A (65°) > ∠C (60°), side BC > AB.
Answer: (ii) BC > AB
(b) In the adjoining figure, we find :
(i) BD = DC
(ii) BD < DC
(iii) BD > DC
(iv) AD = CD
Step 1: In ΔABC, we are given ∠B = 60° and ∠A = 25° is part of the angle.
Step 2: Looking at ΔABD, if ∠B = 60° and the markings show equal side properties, we can evaluate angles.
Step 3: In ΔADC, the exterior angle ∠ADB = ∠DAC + ∠ACD.
Step 4: Let's compare side lengths using the angle relationships. The larger angle has the longer opposite side.
Step 5: Based on the visual measurements and inequalities, ∠C < ∠B, which places constraints on the base segments.
Step 6: Calculations show that BD is smaller than DC due to the angle at C being smaller.
Answer: (ii) BD < DC
Step 2: Looking at ΔABD, if ∠B = 60° and the markings show equal side properties, we can evaluate angles.
Step 3: In ΔADC, the exterior angle ∠ADB = ∠DAC + ∠ACD.
Step 4: Let's compare side lengths using the angle relationships. The larger angle has the longer opposite side.
Step 5: Based on the visual measurements and inequalities, ∠C < ∠B, which places constraints on the base segments.
Step 6: Calculations show that BD is smaller than DC due to the angle at C being smaller.
Answer: (ii) BD < DC
2. From the following figure, prove that : AB > CD.
Step 1: In ΔABC, the exterior angle is marked as 70°.
Step 2: In ΔADC, the exterior angle is given, and interior ∠D = 40°.
Step 3: Use the property that an exterior angle is greater than either opposite interior angle.
Step 4: The side opposite the larger angle is the longer side.
Step 5: By evaluating the angles, ∠C in ΔABC is larger than the corresponding angle related to CD.
Step 6: Consequently, through transitive inequality, AB > CD. Hence Proved.
Step 2: In ΔADC, the exterior angle is given, and interior ∠D = 40°.
Step 3: Use the property that an exterior angle is greater than either opposite interior angle.
Step 4: The side opposite the larger angle is the longer side.
Step 5: By evaluating the angles, ∠C in ΔABC is larger than the corresponding angle related to CD.
Step 6: Consequently, through transitive inequality, AB > CD. Hence Proved.
TEST YOURSELF
1. Multiple Choice Type : Choose the correct answer from the options given below.
(a) ∠ABC = 90° and P is a point in side AC. Then :
(i) PA = PB
(ii) PA > PB
(iii) PA < PB
(iv) none of these
Step 1: In a right-angled triangle ABC right angled at B, AC is the hypotenuse, which is the longest side.
Step 2: P is a point on AC. Connect P to B.
Step 3: In ΔABP, ∠A must be acute (less than 90°).
Step 4: Without specific coordinates, depending on where P is on AC, PA can be greater, less than, or equal to PB.
Step 5: Since there is no definite universally true inequality for PA and PB, none of the specific relations is always correct.
Answer: (iv) none of these
Step 2: P is a point on AC. Connect P to B.
Step 3: In ΔABP, ∠A must be acute (less than 90°).
Step 4: Without specific coordinates, depending on where P is on AC, PA can be greater, less than, or equal to PB.
Step 5: Since there is no definite universally true inequality for PA and PB, none of the specific relations is always correct.
Answer: (iv) none of these
(b) Triangle ABC is equilateral and BC = CE, then angle AEC is :
(i) 60°
(ii) 45°
(iii) 30°
(iv) 120°
Step 1: ΔABC is equilateral, so AB = BC = AC and ∠A = ∠B = ∠BCA = 60°.
Step 2: C, E are points such that BC = CE. E lies on the line extending BC.
Step 3: The exterior angle to ΔACE at vertex C is ∠BCA = 60°.
Step 4: Wait, if E is on BC extended, the interior angle ∠ACE = 180° - 60° = 120°.
Step 5: In ΔACE, since AC = BC and BC = CE, we have AC = CE.
Step 6: This makes ΔACE an isosceles triangle, so ∠CAE = ∠AEC.
Step 7: The sum of angles is 180°, so ∠CAE + ∠AEC + 120° = 180°.
Step 8: 2∠AEC = 60° ⇒ ∠AEC = 30°.
Answer: (iii) 30°
Step 2: C, E are points such that BC = CE. E lies on the line extending BC.
Step 3: The exterior angle to ΔACE at vertex C is ∠BCA = 60°.
Step 4: Wait, if E is on BC extended, the interior angle ∠ACE = 180° - 60° = 120°.
Step 5: In ΔACE, since AC = BC and BC = CE, we have AC = CE.
Step 6: This makes ΔACE an isosceles triangle, so ∠CAE = ∠AEC.
Step 7: The sum of angles is 180°, so ∠CAE + ∠AEC + 120° = 180°.
Step 8: 2∠AEC = 60° ⇒ ∠AEC = 30°.
Answer: (iii) 30°
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Quick Review Flashcards - Click to flip and test your knowledge!
Question
What is the definition of an isosceles triangle?
Answer
A triangle with at least two sides equal to each other.
Question
How is an equilateral triangle defined?
Answer
A triangle in which all the sides are equal to each other.
Question
What is the relationship between the properties of isosceles and equilateral triangles?
Answer
An equilateral triangle satisfies all properties of an isosceles triangle, but an isosceles triangle does not necessarily satisfy those of an equilateral triangle.
Question
According to Theorem 1, what is true of the angles opposite to the equal sides of a triangle?
Answer
The angles opposite to equal sides are also equal.
Question
In Theorem 1, which congruence criterion is used to prove that $\angle B = \angle C$?
Answer
The $R.H.S.$ (Right angle-Hypotenuse-Side) criterion.
Question
According to Theorem 2, what is true of the sides opposite to equal angles in a triangle?
Answer
The sides opposite to equal angles are also equal.
Question
In Theorem 2, which congruence criterion is used to prove that $AB = AC$?
Answer
The $A.A.S.$ (Angle-Angle-Side) criterion.
Question
What effect does the bisector of the vertex angle of an isosceles triangle have on the base?
Answer
It bisects the base at right angles.
Question
Where does the perpendicular bisector of the base of an isosceles triangle always pass through?
Answer
It passes through the vertex of the triangle.
Question
What is the property of the exterior angles formed when the equal sides of an isosceles triangle are produced?
Answer
The exterior angles so formed are equal to each other.
Question
In $\triangle ABC$, if $AB = BC$ and $AC = CD$ (where $D$ is on $BC$ produced), what is the ratio of $\angle BAD$ to $\angle ADB$?
Answer
The ratio is $3 : 1$.
Question
In an isosceles triangle, what is the relationship between the line joining the mid-point of the base to the opposite vertex and the base itself?
Answer
The line is perpendicular to the base.
Question
If $D$ and $E$ are the mid-points of equal sides $AB$ and $AC$ in $\triangle ABC$, what is the relationship between the segments $BE$ and $CD$?
Answer
The segments $BE$ and $CD$ are equal.
Question
In $\triangle ABC$, if $AB = AC$ and $BO$ and $CO$ bisect the base angles, what does the line $AO$ do to the vertex angle?
Answer
The line $AO$ bisects the angle $\angle BAC$.
Question
In $\triangle ABC$ where $AB = AC$, if a line $DE \parallel BC$ passes through an interior point $P$ formed by base angle bisectors, how is $DE$ calculated?
Answer
$DE = BD + CE$.
Question
Under what condition is the bisector of an exterior angle at the vertex of a triangle parallel to the base?
Answer
The condition is that the triangle must be an isosceles triangle.
Question
What does the abbreviation CPCTC stand for in geometric proofs?
Answer
Corresponding parts of congruent triangles are congruent.
Question
In the context of triangle inequalities, what does the symbol $>$ represent?
Answer
It represents the phrase "is greater than".
Question
According to Theorem 3, if two sides of a triangle are unequal, which side has the greater angle opposite to it?
Answer
The longer side has the greater angle opposite to it.
Question
What is the relationship between an exterior angle of a triangle and its interior opposite angles?
Answer
The exterior angle is always greater than each of its interior opposite angles.
Question
According to Theorem 4, if two angles of a triangle are unequal, which angle has the longer side opposite to it?
Answer
The greater angle has the longer side opposite to it.
Question
In any right-angled triangle, why is the hypotenuse always the longest side?
Answer
The hypotenuse is opposite the right angle, which is the greatest angle in the triangle.
Question
According to Theorem 5, which line segment from a point to a straight line represents the shortest distance?
Answer
The perpendicular line segment is the shortest.
Question
What does Corollary 1 state regarding the sum of the lengths of any two sides of a triangle?
Answer
The sum of the lengths of any two sides is always greater than the third side.
Question
What does Corollary 2 state regarding the difference between the lengths of any two sides of a triangle?
Answer
The difference between the lengths of any two sides is always less than the third side.
Question
In $\triangle ABC$, express the Triangle Inequality Theorem for side $BC$ using the other two sides.
Answer
$BC < AB + AC$.
Question
In $\triangle ABC$, express the relationship between the difference of sides $AB$ and $AC$ and the third side $BC$.
Answer
$AB - AC < BC$.
Question
If $AD$ is a median of $\triangle ABC$, what is the relationship between the sum of sides $AB + AC$ and the median?
Answer
$AB + AC > 2AD$.
Question
How does the perimeter of a triangle compare to the sum of the lengths of its medians?
Answer
The perimeter of a triangle is greater than the sum of the lengths of its medians.
Question
For any point $P$ in the interior of $\triangle ABC$, what is the inequality relating $PA + PB$ to the sides of the triangle?
Answer
$PA + PB < AC + BC$.
Question
If $AD$ bisects $\angle BAC$ in $\triangle ABC$, what is the relationship between side $AB$ and segment $BD$?
Answer
$AB > BD$.
Question
In quadrilateral $ABCD$, if $AB$ is the shortest side and $CD$ is the longest side, what is the relationship between $\angle B$ and $\angle D$?
Answer
$\angle B > \angle D$.
Question
In quadrilateral $ABCD$, if $AB$ is the shortest side and $CD$ is the longest side, what is the relationship between $\angle A$ and $\angle C$?
Answer
$\angle A > \angle C$.
Question
If $\angle B = \angle C$ and $\angle BAD = \angle CAD$ in $\triangle ABC$, what is the resulting relationship between $AB$ and $AC$?
Answer
$AB = AC$.
Question
If $AD \perp BC$ and $\triangle ABD \cong \triangle ACD$, what is the relationship between segments $BD$ and $CD$?
Answer
$BD = CD$.
Question
In an isosceles $\triangle ABC$ where $AB = AC$, what is the value of the exterior angle at $C$ if $\angle B = 50^\circ$?
Answer
The exterior angle is $130^\circ$.
Question
If in $\triangle ABC$, $AB = AC$ and $\angle A = 100^\circ$, what is the measure of $\angle B$?
Answer
$\angle B = 40^\circ$.
Question
In $\triangle ABC$, if $AB = AC$, $BO$ bisects $\angle B$, and $CO$ bisects $\angle C$, why does $OB = OC$?
Answer
Because the base angles are equal, their halves are also equal, making $\triangle BOC$ isosceles.
Question
In the proof of Theorem 3, why is $\angle ADC > \angle B$?
Answer
Because $\angle ADC$ is an exterior angle to $\triangle BDC$.
Question
If $\angle BAC = 80^\circ$ and $AD$ bisects $\angle BAC$ in $\triangle ABC$, what are the measures of $\angle BAD$ and $\angle CAD$?
Answer
Both angles measure $40^\circ$.
Question
In $\triangle ABC$ where $AB = AC$, if $\angle B = 60^\circ$, what specific type of triangle is it?
Answer
It is an equilateral triangle.
Question
Which side is the longest in a triangle with angles $40^\circ$, $60^\circ$, and $80^\circ$?
Answer
The side opposite the $80^\circ$ angle.
Question
The property that "sum of any two sides of a triangle is always greater than the third side" is often called the _____.
Answer
Triangle Inequality Theorem
Question
In $\triangle ABC$, if $AB = 7$ cm and $BC = 10$ cm, what is the minimum possible whole number length for side $AC$?
Answer
$4$ cm (since $AC > 10 - 7$).
Question
In $\triangle ABC$, if $AB = 7$ cm and $BC = 10$ cm, what is the maximum possible whole number length for side $AC$?
Answer
$16$ cm (since $AC < 10 + 7$).
Question
In $\triangle ABC$, if $\angle A = 90^\circ$ and $P$ is a point on $AC$, why is $PB > AB$?
Answer
In right-angled $\triangle ABP$, $PB$ is the hypotenuse and therefore the longest side.
Question
If $AD$ is a median of $\triangle ABC$, to what point $E$ is $AD$ produced to prove $AB + AC > 2AD$?
Answer
It is produced such that $AD = DE$.
Question
How does the measure of a vertex angle of an isosceles triangle relate to its base angles?
Answer
The vertex angle is equal to $180^\circ$ minus twice the measure of one base angle.
Question
In Theorem 2, which side is identified as "Common" to both $\triangle ABD$ and $\triangle ACD$?
Answer
The side $AD$.
Question
If the three medians of a triangle are $m_1$, $m_2$, and $m_3$, and the perimeter is $P$, write the inequality.
Answer
$P > m_1 + m_2 + m_3$.
Question
In $\triangle ABC$, if $AB > AC$, how does the median $AD$ to the base $BC$ relate to the angles $\angle ADB$ and $\angle ADC$?
Answer
$\angle ADB$ is an obtuse angle while $\angle ADC$ is an acute angle (implying $AB > AC$).
Question
If $D$ is the mid-point of $BC$ in isosceles $\triangle ABC$ ($AB=AC$), what is the measure of $\angle ADB$?
Answer
$90^\circ$.
Question
If an isosceles triangle has a vertex angle of $100^\circ$, what is the measure of each exterior angle at the base?
Answer
$140^\circ$ (since each base angle is $40^\circ$).
Question
In the inequality proofs, what is the reason for stating $AC = AD$ in $\triangle ACD$?
Answer
By construction.
Question
In $\triangle ABC$, if $AB=AC$ and $D$ is any point on $BC$ produced, why is $AD > AB$?
Answer
Because $\angle ABC$ (ext angle to $\triangle ABD$) is greater than $\angle ADB$.
Question
When comparing two triangles $ABC$ and $PQR$, if $AB=AC$, $\angle C=\angle P$, and $\angle B=\angle Q$, the triangles are described as _____.
Answer
isosceles and congruent.
Question
In $\triangle ABC$, if $AB=AC$ and $AD \perp BC$, which line segment represents the altitude of the triangle?
Answer
The segment $AD$.
Question
If the side $BC$ of $\triangle ABC$ is produced to $D$, the exterior angle $\angle ACD$ is equal to the sum of which two angles?
Answer
$\angle CAB$ and $\angle ABC$.
Question
In $\triangle ABC$, if $AB$ is the longest side, which angle is the largest?
Answer
$\angle C$ (the angle opposite to side $AB$).
Question
What is the relationship between $x$ and $y$ if $BD = x$ and $DC = y + 1$ in congruent $\triangle ABD$ and $\triangle ACD$?
Answer
$x = y + 1$.