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TRIANGLES - Q&A

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD. What can you say about BC and BD?

Given:
In quadrilateral ACBD, AC = AD.
AB bisects ∠ A, which means ∠ CAB = ∠ DAB.

To Prove: Δ ABC ≅ Δ ABD.

Proof:
In Δ ABC and Δ ABD,
1. AC = AD (Given)
2. ∠ CAB = ∠ DAB (Since AB bisects ∠ A)
3. AB = AB (Common side)

Therefore, by SAS (Side-Angle-Side) congruence rule,
Δ ABC ≅ Δ ABD.

Since the triangles are congruent, their corresponding parts are equal (CPCT - Corresponding Parts of Congruent Triangles).
Therefore, BC = BD.

2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that

(i) Δ ABD ≅ Δ BAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC

Given: AD = BC and ∠ DAB = ∠ CBA.

(i) Proof for Δ ABD ≅ Δ BAC:
In Δ ABD and Δ BAC,
1. AD = BC (Given)
2. ∠ DAB = ∠ CBA (Given)
3. AB = AB (Common side)
So, by SAS congruence rule,
Δ ABD ≅ Δ BAC.

(ii) Proof for BD = AC:
Since Δ ABD ≅ Δ BAC, their corresponding sides are equal.
Therefore, BD = AC (by CPCT).

(iii) Proof for ∠ ABD = ∠ BAC:
Since Δ ABD ≅ Δ BAC, their corresponding angles are equal.
Therefore, ∠ ABD = ∠ BAC (by CPCT).

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Given:
AD and BC are perpendicular to AB, so ∠ OAD = 90° and ∠ OBC = 90°.
AD = BC.

To Prove: CD bisects AB (i.e., OA = OB).

Proof:
In Δ OBC and Δ OAD,
1. ∠ OBC = ∠ OAD (Each 90°)
2. ∠ BOC = ∠ AOD (Vertically opposite angles)
3. BC = AD (Given)

So, by AAS (Angle-Angle-Side) congruence rule,
Δ OBC ≅ Δ OAD.

Since the triangles are congruent, their corresponding parts are equal.
OB = OA (by CPCT).
Since O is the midpoint of AB, CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that Δ ABC ≅ Δ CDA.

Given: l || m and p || q.
To Prove: Δ ABC ≅ Δ CDA.

Proof:
Since l || m and AC is a transversal, alternate interior angles are equal.
So, ∠ BCA = ∠ DAC ... (i)

Since p || q and AC is a transversal, alternate interior angles are equal.
So, ∠ BAC = ∠ DCA ... (ii)

In Δ ABC and Δ CDA,
1. ∠ BCA = ∠ DAC (from i)
2. AC = CA (Common side)
3. ∠ BAC = ∠ DCA (from ii)

Therefore, by ASA (Angle-Side-Angle) congruence rule,
Δ ABC ≅ Δ CDA.

5. Line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Fig. 7.20). Show that:

(i) Δ APB ≅ Δ AQB

(ii) BP = BQ or B is equidistant from the arms of ∠ A.

Given:
Line l bisects ∠ A, so ∠ PAB = ∠ QAB.
BP ⊥ AP and BQ ⊥ AQ, so ∠ APB = 90° and ∠ AQB = 90°.

(i) Proof for Δ APB ≅ Δ AQB:
In Δ APB and Δ AQB,
1. ∠ APB = ∠ AQB (Each 90°)
2. ∠ PAB = ∠ QAB (Given)
3. AB = AB (Common side)
So, by AAS congruence rule,
Δ APB ≅ Δ AQB.

(ii) Proof for BP = BQ:
Since Δ APB ≅ Δ AQB,
BP = BQ (by CPCT).
This means B is equidistant from the arms of ∠ A.

6. In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.

Given: AC = AE, AB = AD, and ∠ BAD = ∠ EAC.
To Prove: BC = DE.

Proof:
We are given ∠ BAD = ∠ EAC.
Add ∠ DAC to both sides:
∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC
∠ BAC = ∠ DAE ... (i)

Now, in Δ BAC and Δ DAE,
1. AB = AD (Given)
2. ∠ BAC = ∠ DAE (from i)
3. AC = AE (Given)

Therefore, by SAS congruence rule,
Δ BAC ≅ Δ DAE.

So, BC = DE (by CPCT).

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig. 7.22). Show that

(i) Δ DAP ≅ Δ EBP

(ii) AD = BE

Given:
P is the mid-point of AB, so AP = BP.
∠ BAD = ∠ ABE.
∠ EPA = ∠ DPB.

(i) Proof for Δ DAP ≅ Δ EBP:
We are given ∠ EPA = ∠ DPB.
Add ∠ EPD to both sides:
∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD
∠ APD = ∠ BPE ... (i)

In Δ DAP and Δ EBP,
1. ∠ DAP = ∠ EBP (Given, same as ∠ BAD = ∠ ABE)
2. AP = BP (P is mid-point)
3. ∠ APD = ∠ BPE (from i)
So, by ASA congruence rule,
Δ DAP ≅ Δ EBP.

(ii) Proof for AD = BE:
Since Δ DAP ≅ Δ EBP,
AD = BE (by CPCT).

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:

(i) Δ AMC ≅ Δ BMD

(ii) ∠ DBC is a right angle.

(iii) Δ DBC ≅ Δ ACB

(iv) CM = 1/2 AB

Given: ∠ C = 90°, M is mid-point of AB (AM = BM), and DM = CM.

(i) In Δ AMC and Δ BMD,
1. AM = BM (Given)
2. ∠ AMC = ∠ BMD (Vertically opposite angles)
3. CM = DM (Given)
So, Δ AMC ≅ Δ BMD (by SAS).

(ii) Since Δ AMC ≅ Δ BMD, ∠ MAC = ∠ MBD (by CPCT).
These are alternate interior angles for lines AC and BD intersected by AB.
Therefore, AC || BD.
Since AC || BD and BC is a transversal, consecutive interior angles sum to 180°.
∠ DBC + ∠ ACB = 180°
∠ DBC + 90° = 180°
∠ DBC = 90°.

(iii) In Δ DBC and Δ ACB,
1. DB = AC (Since Δ AMC ≅ Δ BMD, CPCT gives AC = BD)
2. ∠ DBC = ∠ ACB (Each 90°)
3. BC = BC (Common)
So, Δ DBC ≅ Δ ACB (by SAS).

(iv) Since Δ DBC ≅ Δ ACB, DC = AB (by CPCT).
We know DM = CM, so DC = 2CM.
Therefore, 2CM = AB
CM = 1/2 AB.


Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O. Join A to O. Show that :

(i) OB = OC

(ii) AO bisects ∠ A

(i) Given AB = AC.
So, ∠ C = ∠ B (Angles opposite to equal sides are equal).
Since OB and OC are bisectors, ∠ OBC = 1/2 ∠ B and ∠ OCB = 1/2 ∠ C.
Since ∠ B = ∠ C, 1/2 ∠ B = 1/2 ∠ C.
So, ∠ OBC = ∠ OCB.
In Δ OBC, sides opposite to equal angles are equal.
Therefore, OB = OC.

(ii) In Δ AOB and Δ AOC,
1. AB = AC (Given)
2. AO = AO (Common)
3. OB = OC (Proved above)
So, Δ AOB ≅ Δ AOC (by SSS).
Therefore, ∠ BAO = ∠ CAO (by CPCT).
Hence, AO bisects ∠ A.

2. In Δ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Δ ABC is an isosceles triangle in which AB = AC.

Given: AD ⊥ BC (so ∠ ADB = ∠ ADC = 90°) and BD = CD (Bisector).
To Prove: AB = AC.

Proof:
In Δ ABD and Δ ACD,
1. BD = CD (Given)
2. ∠ ADB = ∠ ADC (Each 90°)
3. AD = AD (Common)
So, Δ ABD ≅ Δ ACD (by SAS).
Therefore, AB = AC (by CPCT).
Since two sides are equal, Δ ABC is an isosceles triangle.

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Given: AB = AC, BE ⊥ AC, CF ⊥ AB.
To Prove: BE = CF.

Proof:
In Δ ABE and Δ ACF,
1. ∠ A = ∠ A (Common)
2. ∠ AEB = ∠ AFC (Each 90°)
3. AB = AC (Given)
So, Δ ABE ≅ Δ ACF (by AAS).
Therefore, BE = CF (by CPCT).

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Given: BE = CF, BE ⊥ AC, CF ⊥ AB.

(i) In Δ ABE and Δ ACF,
1. ∠ A = ∠ A (Common)
2. ∠ AEB = ∠ AFC (Each 90°)
3. BE = CF (Given)
So, Δ ABE ≅ Δ ACF (by AAS).

(ii) Since Δ ABE ≅ Δ ACF,
AB = AC (by CPCT).
Therefore, ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ ABD = ∠ ACD.

Given: AB = AC (for Δ ABC) and DB = DC (for Δ DBC).
To Prove: ∠ ABD = ∠ ACD.

Proof:
In Δ ABC, since AB = AC, ∠ ABC = ∠ ACB (Angles opposite to equal sides are equal) ... (i)
In Δ DBC, since DB = DC, ∠ DBC = ∠ DCB ... (ii)

Adding (i) and (ii):
∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
∠ ABD = ∠ ACD.
Hence Proved.

6. Δ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.

Given: AB = AC and AD = AB. So, AC = AD.
To Prove: ∠ BCD = 90°.

Proof:
In Δ ABC, AB = AC. So, ∠ ACB = ∠ ABC. Let this be x.
In Δ ACD, AC = AD. So, ∠ ACD = ∠ ADC. Let this be y.

In Δ BCD,
Sum of angles = 180°
∠ B + ∠ C + ∠ D = 180°
x + (x + y) + y = 180°
2x + 2y = 180°
2(x + y) = 180°
x + y = 90°
Since ∠ BCD = x + y,
∠ BCD = 90°.

7. ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.

Given: ∠ A = 90°, AB = AC.
Since AB = AC, ∠ C = ∠ B (Angles opposite to equal sides).
In Δ ABC,
∠ A + ∠ B + ∠ C = 180°
90° + ∠ B + ∠ B = 180°
2 ∠ B = 90°
∠ B = 45°.
Since ∠ C = ∠ B, ∠ C = 45°.
Answer: ∠ B = 45°, ∠ C = 45°.

8. Show that the angles of an equilateral triangle are 60° each.

Let ABC be an equilateral triangle, so AB = BC = AC.
Since AB = AC, ∠ C = ∠ B.
Since BC = AC, ∠ A = ∠ B.
So, ∠ A = ∠ B = ∠ C.

Sum of angles = 180°
∠ A + ∠ B + ∠ C = 180°
3 ∠ A = 180°
∠ A = 60°.
Therefore, ∠ A = ∠ B = ∠ C = 60°.


Exercise 7.3

1. Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that

(i) Δ ABD ≅ Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC.

Given: AB = AC (Isosceles Δ ABC), DB = DC (Isosceles Δ DBC).

(i) In Δ ABD and Δ ACD,
1. AB = AC (Given)
2. DB = DC (Given)
3. AD = AD (Common)
So, Δ ABD ≅ Δ ACD (by SSS).

(ii) From (i), ∠ BAP = ∠ CAP (by CPCT).
In Δ ABP and Δ ACP,
1. AB = AC (Given)
2. ∠ BAP = ∠ CAP (Proved)
3. AP = AP (Common)
So, Δ ABP ≅ Δ ACP (by SAS).

(iii) From (i), ∠ BAD = ∠ CAD, so AP bisects ∠ A.
In Δ BDP and Δ CDP,
1. DB = DC (Given)
2. BP = CP (Since Δ ABP ≅ Δ ACP, CPCT)
3. DP = DP (Common)
So, Δ BDP ≅ Δ CDP (by SSS).
Therefore, ∠ BDP = ∠ CDP (by CPCT).
So, AP bisects ∠ D as well.

(iv) Since Δ ABP ≅ Δ ACP, BP = CP.
Also, ∠ APB = ∠ APC (by CPCT).
But ∠ APB + ∠ APC = 180° (Linear Pair).
2 ∠ APB = 180° ⇒ ∠ APB = 90°.
Since BP = CP and AP ⊥ BC, AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠ A.

Given: AB = AC, AD ⊥ BC (∠ ADB = ∠ ADC = 90°).

(i) In right triangles Δ ADB and Δ ADC,
1. Hypotenuse AB = Hypotenuse AC (Given)
2. Side AD = Side AD (Common)
So, Δ ADB ≅ Δ ADC (by RHS).
Therefore, BD = CD (by CPCT).
So, AD bisects BC.

(ii) Since Δ ADB ≅ Δ ADC,
∠ BAD = ∠ CAD (by CPCT).
So, AD bisects ∠ A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40). Show that:

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Given: AB = PQ, BC = QR, AM = PN.
Since AM and PN are medians, BM = 1/2 BC and QN = 1/2 QR.
Since BC = QR, BM = QN.

(i) In Δ ABM and Δ PQN,
1. AB = PQ (Given)
2. AM = PN (Given)
3. BM = QN (Proved)
So, Δ ABM ≅ Δ PQN (by SSS).

(ii) Since Δ ABM ≅ Δ PQN, ∠ B = ∠ Q (by CPCT).
In Δ ABC and Δ PQR,
1. AB = PQ (Given)
2. ∠ B = ∠ Q (Proved)
3. BC = QR (Given)
So, Δ ABC ≅ Δ PQR (by SAS).

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Given: BE = CF, BE ⊥ AC, CF ⊥ AB.
To Prove: AB = AC.

Proof:
In right triangles Δ BEC and Δ CFB,
1. Hypotenuse BC = Hypotenuse BC (Common)
2. Side BE = Side CF (Given)
So, Δ BEC ≅ Δ CFB (by RHS).

Therefore, ∠ BCE = ∠ CBF (by CPCT).
This means ∠ C = ∠ B.
In Δ ABC, since ∠ B = ∠ C, sides opposite to equal angles are equal.
So, AC = AB.
Hence, Δ ABC is isosceles.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.

Given: AB = AC, AP ⊥ BC (∠ APB = ∠ APC = 90°).
To Prove: ∠ B = ∠ C.

Proof:
In right triangles Δ ABP and Δ ACP,
1. Hypotenuse AB = Hypotenuse AC (Given)
2. Side AP = Side AP (Common)
So, Δ ABP ≅ Δ ACP (by RHS).

Therefore, ∠ B = ∠ C (by CPCT).

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Quick Review Flashcards - Click to flip and test your knowledge!
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What is the geometric definition of a triangle?
Answer
A closed figure formed by three intersecting lines.
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Answer
Three sides, three angles, and three vertices.
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Answer
Equal in all respects, meaning figures have both the same shape and the same size.
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Their radii must be equal.
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The measures of their sides must be equal.
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Answer
When the sides and angles of one triangle are equal to the corresponding sides and angles of the other.
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Answer
The symbol $\cong$.
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Answer
A one-to-one correspondence between vertices.
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Answer
The order indicates which specific vertices and sides correspond to each other.
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What does the acronym 'CPCT' stand for?
Answer
Corresponding parts of congruent triangles.
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According to the SAS congruence rule, what components must be equal in two triangles?
Answer
Two sides and the included angle.
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In the SAS congruence rule, what is meant by the 'included angle'?
Answer
The angle located between the two sides that are being compared for equality.
Question
How is the SAS congruence rule categorized in this text regarding its provability?
Answer
It is accepted as an axiom (Axiom 7.1) and cannot be proved using previously known results.
Question
The _____ rule is not a valid criterion for congruence because the equal angle must be included between the equal sides.
Answer
SSA or ASS
Question
State the Angle-Side-Angle (ASA) congruence rule.
Answer
Two triangles are congruent if two angles and the included side of one are equal to the corresponding parts of the other.
Question
Which axiom is utilized as the basis to prove the ASA congruence theorem?
Answer
The SAS congruence axiom.
Question
What is the core requirement of the AAS congruence rule?
Answer
Two pairs of angles and one pair of corresponding sides must be equal.
Question
Why is the AAS rule valid if the ASA rule is already established?
Answer
The sum of angles in a triangle is always $180^\circ$, ensuring the third pair of angles is also equal.
Question
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Answer
Triangles with the same angles can have different side lengths, resulting in different sizes.
Question
What is the Side-Side-Side (SSS) congruence rule?
Answer
Two triangles are congruent if all three sides of one triangle are equal to the three sides of the other.
Question
What are the specific requirements for the RHS congruence rule?
Answer
The hypotenuse and one side of two right triangles must be equal.
Question
In the acronym RHS, what do the letters represent?
Answer
Right angle, Hypotenuse, and Side.
Question
What defines an isosceles triangle?
Answer
A triangle in which at least two sides are equal.
Question
According to Theorem 7.2, what is true about the angles opposite to the equal sides of an isosceles triangle?
Answer
The angles are equal to each other.
Question
Theorem 7.3 states that the sides opposite to equal angles of a triangle are _____.
Answer
Equal
Question
What is the measure of each interior angle in an equilateral triangle?
Answer
$60^\circ$.
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If line $l$ is the perpendicular bisector of segment $AB$, what can be said about any point $P$ on $l$ relative to $A$ and $B$?
Answer
Point $P$ is equidistant from points $A$ and $B$.
Question
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Answer
They are vertically opposite angles.
Question
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Answer
Alternate interior angles are equal.
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Answer
It bisects side $BC$, meaning $BM = MC$.
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Answer
A line segment from a vertex perpendicular to the opposite side.
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Answer
The SAS rule.
Question
In right triangle proofs, if the hypotenuse and one leg are equal, which congruence rule is most direct?
Answer
The RHS rule.
Question
If two triangles are congruent, their corresponding areas must be _____.
Answer
Equal
Question
What construction is used in the textbook to prove Theorem 7.2 regarding isosceles triangle angles?
Answer
Drawing the bisector of the vertex angle.
Question
How can one prove that the sides opposite to equal angles are equal (Theorem 7.3)?
Answer
By using the ASA congruence rule.
Question
What is the term for a triangle where all three sides are equal?
Answer
Equilateral triangle.
Question
In Exercise 7.1, what is concluded about segments $BC$ and $BD$ if $\Delta ABC \cong \Delta ABD$?
Answer
They are equal due to CPCT.
Question
What does it mean if a line $CD$ bisects segment $AB$?
Answer
It passes through the mid-point of $AB$, dividing it into two equal segments.
Question
If point $P$ is equidistant from two intersecting lines, what does the line $AP$ do to the angle between those lines?
Answer
It bisects the angle.