MID-POINT THEOREM AND ITS CONVERSE [Including Intercept Theorem] - Questions & Answers

EXERCISE 11(A)

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In the given figure, ABCD is a rectangle. As per the given information, the length of PQ is :
(i) 12 cm
(ii) 14 cm
(iii) 20 cm
(iv) 10 cm
Step 1: The figure shows rectangle ABCD with sides AD = 12 cm and DC = 16 cm.
Step 2: Tick marks show P and Q are mid-points of AB and BC respectively.
Step 3: In right-angled triangle ABC, using Pythagoras theorem, AC = √(AB² + BC²) = √(16² + 12²) = √(256 + 144) = 20 cm.
Step 4: By the mid-point theorem, the line segment joining the mid-points of two sides is half of the third side.
Step 5: Therefore, PQ = 1/2 × AC = 1/2 × 20 = 10 cm.
Answer: (iv) 10 cm

(b) The quadrilateral obtained by joining the mid-points (in order) of the sides of quadrilateral ABCD is :
(i) rectangle
(ii) rhombus
(iii) parallelogram
(iv) square
Step 1: By joining the mid-points of adjacent sides of any quadrilateral, the opposite sides of the resulting figure are parallel to the same diagonal.
Step 2: Because they are parallel to the same diagonal, they are parallel to each other, forming a parallelogram.
Answer: (iii) parallelogram

(c) If BC = 12 cm, AB = 14·8 cm, AC = 12·8 cm, the perimeter of quadrilateral BCYX is :
(i) 31·8 cm
(ii) 15·9 cm
(iii) 29·8 cm
(iv) 32·8 cm
Step 1: The tick marks show X and Y are mid-points of AB and AC respectively.
Step 2: BX = 1/2 × AB = 1/2 × 14.8 = 7.4 cm.
Step 3: CY = 1/2 × AC = 1/2 × 12.8 = 6.4 cm.
Step 4: By mid-point theorem, XY = 1/2 × BC = 1/2 × 12 = 6 cm.
Step 5: Perimeter of BCYX = BC + CY + XY + BX = 12 + 6.4 + 6 + 7.4 = 31.8 cm.
Answer: (i) 31.8 cm

(d) In the given figure, AB = AC. P, Q and R are mid-points of sides BC, CA and AB respectively, then triangle PQR is :
(i) scalene
(ii) isosceles
(iii) equilateral
(iv) obtuse angled
Step 1: Since P, Q, and R are mid-points, by the mid-point theorem, PQ = 1/2 AB and PR = 1/2 AC.
Step 2: It is given that AB = AC.
Step 3: Therefore, 1/2 AB = 1/2 AC, which means PQ = PR.
Step 4: Since two sides of triangle PQR are equal, it is an isosceles triangle.
Answer: (ii) isosceles

(e) P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively of rectangle ABCD, then quadrilateral PQRS is :
(i) rectangle
(ii) rhombus
(iii) square
(iv) parallelogram
Step 1: The diagonals of a rectangle are equal in length, so AC = BD.
Step 2: By the mid-point theorem, PQ = SR = 1/2 AC.
Step 3: Similarly, PS = QR = 1/2 BD.
Step 4: Since AC = BD, all four sides are equal: PQ = QR = RS = SP.
Step 5: A parallelogram with all sides equal is a rhombus.
Answer: (ii) rhombus

2. In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC at N. Find the lengths of AN and MN, if BC = 7 cm and AC = 5 cm.
Step 1: In triangle ABC, M is the mid-point of AB and MN is parallel to BC.
Step 2: By the converse of the mid-point theorem, N must be the mid-point of AC.
Step 3: Therefore, AN = 1/2 × AC = 1/2 × 5 = 2.5 cm.
Step 4: By the mid-point theorem, MN = 1/2 × BC.
Step 5: Therefore, MN = 1/2 × 7 = 3.5 cm.
Answer: AN = 2.5 cm, MN = 3.5 cm

3. Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
Step 1: Let ABCD be a rectangle with diagonals AC and BD.
Step 2: Let P, Q, R, S be the mid-points of AB, BC, CD, and DA respectively.
Step 3: We know that the diagonals of a rectangle are equal, so AC = BD.
Step 4: In triangle ABC, P and Q are mid-points of AB and BC, so PQ = 1/2 AC and PQ // AC.
Step 5: In triangle ADC, S and R are mid-points of AD and DC, so SR = 1/2 AC and SR // AC.
Step 6: Therefore, PQ = SR and PQ // SR, making PQRS a parallelogram.
Step 7: Similarly, using triangle ABD and BCD, PS = 1/2 BD and QR = 1/2 BD.
Step 8: Since AC = BD, it follows that PQ = QR = RS = SP.
Step 9: Since all four sides of parallelogram PQRS are equal, it is a rhombus. Hence Proved.

4. D, E and F are the mid-points of the sides AB, BC and CA of an isosceles triangle ABC in which AB = BC. Prove that triangle DEF is also isosceles.
Step 1: In triangle ABC, D and F are the mid-points of AB and AC respectively.
Step 2: By the mid-point theorem, DF = 1/2 BC.
Step 3: Similarly, E and F are mid-points of BC and AC, so EF = 1/2 AB.
Step 4: It is given that triangle ABC is isosceles with AB = BC.
Step 5: Substituting AB with BC in step 3 gives EF = 1/2 BC.
Step 6: From steps 2 and 5, DF = EF.
Step 7: Since two sides of triangle DEF are equal in length, triangle DEF is an isosceles triangle. Hence Proved.

5. The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that : PR = 1/2 (AB + CD).
Step 1: Join the diagonal BD, and let it intersect the line segment PR at point Q.
Step 2: In triangle ABD, P is the mid-point of AD and PQ // AB.
Step 3: By the converse of the mid-point theorem, Q is the mid-point of BD.
Step 4: By the mid-point theorem, PQ = 1/2 AB.
Step 5: In triangle BCD, Q is the mid-point of BD and QR // DC (since PR // AB and AB // DC).
Step 6: By the converse of the mid-point theorem, R is the mid-point of BC.
Step 7: By the mid-point theorem, QR = 1/2 CD.
Step 8: Adding the two parts together: PR = PQ + QR = 1/2 AB + 1/2 CD.
Step 9: Therefore, PR = 1/2 (AB + CD). Hence Proved.

6. The figure, given below, shows a trapezium ABCD. M and N are the mid-points of the non-parallel sides AD and BC respectively. Find :
(i) MN, if AB = 11 cm and DC = 8 cm.
(ii) AB, if DC = 20 cm and MN = 27 cm.
(iii) DC, if MN = 15 cm and AB = 23 cm.
Step 1: As proved in the previous theorem, the line joining mid-points of non-parallel sides of a trapezium equals half the sum of parallel sides.
Step 2: Formula: MN = 1/2 (AB + DC).
Step 3: For (i): MN = 1/2 (11 + 8) = 1/2 (19) = 9.5 cm.
Step 4: For (ii): 27 = 1/2 (AB + 20) → 54 = AB + 20 → AB = 34 cm.
Step 5: For (iii): 15 = 1/2 (23 + DC) → 30 = 23 + DC → DC = 7 cm.
Answer: (i) 9.5 cm, (ii) 34 cm, (iii) 7 cm

7. The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.
Step 1: Let the quadrilateral be ABCD with diagonals AC and BD intersecting at 90 degrees.
Step 2: Let P, Q, R, S be the mid-points of AB, BC, CD, and DA respectively.
Step 3: By mid-point theorem, PQ // AC and SR // AC. Thus PQ // SR.
Step 4: Similarly, PS // BD and QR // BD. Thus PS // QR.
Step 5: Since opposite sides are parallel, PQRS is a parallelogram.
Step 6: We are given that diagonals AC and BD are perpendicular to each other.
Step 7: Since PQ is parallel to AC and QR is parallel to BD, the angle between PQ and QR must be the same as the angle between AC and BD.
Step 8: Therefore, the angle between PQ and QR is 90 degrees.
Step 9: A parallelogram with an interior angle of 90 degrees is a rectangle. Hence Proved.

8. L and M are the mid-points of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
Step 1: In parallelogram ABCD, AB is parallel and equal to DC.
Step 2: Since L and M are mid-points, AL = 1/2 AB and MC = 1/2 DC.
Step 3: Therefore, AL is parallel and equal to MC, making ALCM a parallelogram.
Step 4: Because ALCM is a parallelogram, AM // LC. This implies that DL // BM.
Step 5: Let DL and BM intersect diagonal AC at points P and Q respectively.
Step 6: In triangle ABQ, L is the mid-point of AB and LP is parallel to BQ (since DL // BM).
Step 7: By converse of the mid-point theorem, P is the mid-point of AQ, meaning AP = PQ.
Step 8: In triangle CDP, M is the mid-point of CD and MQ is parallel to DP (since AM // LC).
Step 9: By converse of the mid-point theorem, Q is the mid-point of CP, meaning PQ = QC.
Step 10: From steps 7 and 9, AP = PQ = QC. Thus, DL and BM trisect diagonal AC. Hence Proved.

9. ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and AC respectively. Prove that EFGH is a rhombus.
Step 1: In triangle ABD, E and F are the mid-points of AB and BD.
Step 2: By the mid-point theorem, EF // AD and EF = 1/2 AD.
Step 3: In triangle ACD, H and G are the mid-points of AC and CD.
Step 4: By the mid-point theorem, HG // AD and HG = 1/2 AD.
Step 5: From steps 2 and 4, EF // HG and EF = HG, which means EFGH is a parallelogram.
Step 6: In triangle ABC, E and H are the mid-points of AB and AC.
Step 7: By the mid-point theorem, EH // BC and EH = 1/2 BC.
Step 8: We are given that AD = BC. Therefore, 1/2 AD = 1/2 BC.
Step 9: Substituting into our equations, we find EF = EH.
Step 10: Since adjacent sides of the parallelogram EFGH are equal, it is a rhombus. Hence Proved.

10. A parallelogram ABCD has P the mid-point of DC and Q a point of AC such that CQ = 1/4 AC. PQ produced meets BC at R. Prove that :
(i) R is the mid-point of BC,
(ii) PR = 1/2 DB.
Step 1: Let the diagonals AC and BD intersect at point O. In a parallelogram, diagonals bisect each other, so CO = 1/2 AC.
Step 2: We are given that CQ = 1/4 AC. This means CQ is exactly half of CO.
Step 3: Therefore, Q is the mid-point of the segment CO.
Step 4: In triangle DCO, P is the mid-point of DC (given) and Q is the mid-point of CO.
Step 5: By the mid-point theorem, PQ is parallel to DO, which means line PR is parallel to diagonal DB.
Step 6: Now, consider triangle BCD.
Step 7: P is the mid-point of DC and line PR is parallel to DB.
Step 8: By the converse of the mid-point theorem, R must be the mid-point of side BC. (Proves i)
Step 9: Also, in triangle BCD, since P and R are mid-points of two sides, by the mid-point theorem, PR = 1/2 DB. (Proves ii). Hence Proved.

EXERCISE 11(B)

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) In the given figure, l // m // n and D is mid-point of CE. If AE = 12·6 cm, then BD is :
(i) 12·6 cm
(ii) 25·2 cm
(iii) 6·3 cm
(iv) 18·9 cm
Step 1: By observing the figure and intercepts, l, m, n are parallel lines cut by transversals AC and CE.
Step 2: Since D is the mid-point of CE, CD = DE, meaning line m makes equal intercepts on transversal CE.
Step 3: By the Equal Intercept Theorem, line m must also make equal intercepts on transversal AC, so B is the mid-point of AC.
Step 4: In triangle ACE, B is the mid-point of AC and D is the mid-point of CE.
Step 5: By the mid-point theorem, BD is parallel to AE and BD = 1/2 AE.
Step 6: BD = 1/2 × 12.6 = 6.3 cm.
Answer: (iii) 6.3 cm

(b) In a trapezium ABCD, AB//DC, E is mid-point of AD and F is mid-point of BC, then :
(i) 2EF = 1/2 (AB + DC)
(ii) 2EF = AB + DC
(iii) EF = AB + DC
(iv) EF = 1/2 × AB × DC
Step 1: The theorem for the line joining the mid-points of the non-parallel sides of a trapezium states it equals half the sum of parallel sides.
Step 2: The equation is EF = 1/2 (AB + DC).
Step 3: Multiplying both sides by 2 gives 2EF = AB + DC.
Answer: (ii) 2EF = AB + DC

(c) The given figure shows a parallelogram ABCD in which E is mid-point of AD and DF//EB. Then, BF is equal to :
(i) AD
(ii) BE
(iii) AE
(iv) AB
Step 1: In the figure, consider triangle ADF formed by sides AD, DF, and the extension of AB to F.
Step 2: We are given that E is the mid-point of AD.
Step 3: It is also given that EB is parallel to DF.
Step 4: In triangle ADF, line EB passes through the mid-point of AD and is parallel to side DF.
Step 5: By the converse of the mid-point theorem, B must be the mid-point of side AF.
Step 6: Because B is the mid-point of AF, the length of BF is equal to the length of AB.
Answer: (iv) AB

(d) In the given figure AD and BE are medians, then ED is equal to :
(i) 2AB
(ii) 1/2 AB
(iii) 1/4 AB
(iv) 1/8 AB
Step 1: In a triangle, a median connects a vertex to the mid-point of the opposite side.
Step 2: Since AD is a median, D is the mid-point of BC.
Step 3: Since BE is a median, E is the mid-point of AC.
Step 4: By the mid-point theorem, the segment connecting the mid-points of two sides of a triangle is half the length of the third side.
Step 5: Therefore, ED = 1/2 AB.
Answer: (ii) 1/2 AB

(e) In the quadrilateral ABCD, if AB//CD, E is mid-point of side AD and F is mid-point of BC. If AB = 20 cm and EF = 16 cm, the length of side DC is :
(i) 18 cm
(ii) 12 cm
(iii) 24 cm
(iv) 32 cm
Step 1: The figure is a trapezium. The length of the mid-segment EF is given by EF = 1/2 (AB + DC).
Step 2: Substitute the known values: 16 = 1/2 (20 + DC).
Step 3: Multiply both sides by 2: 32 = 20 + DC.
Step 4: Subtract 20 from 32: DC = 12 cm.
Answer: (ii) 12 cm

2. Use the following figure to find :
(i) BC, if AB = 7·2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4·1 cm.
(iv) DF, if CG = 11 cm.
Step 1: Based on the figure, lines passing through CG, BDF, and AE are parallel to each other.
Step 2: The tick marks show that CB = BA, meaning the parallel lines make equal intercepts on transversal CA.
Step 3: By the Equal Intercept theorem, they will make equal intercepts on any other transversal.
Step 4: For (i): Since CB = BA, BC = AB = 7.2 cm.
Step 5: For (ii): Transversal GE is intercepted equally, so GF = FE. GE = GF + FE = 4 + 4 = 8 cm.
Step 6: For (iii): In triangle CAE, B and D are mid-points of CA and CE. BD = 1/2 AE. So, AE = 2 × 4.1 = 8.2 cm.
Step 7: For (iv): In triangle CGE, D and F are mid-points of CE and GE. DF = 1/2 CG. So, DF = 1/2 × 11 = 5.5 cm.
Answer: (i) 7.2 cm, (ii) 8 cm, (iii) 8.2 cm, (iv) 5.5 cm

3. In the figure, given below, 2AD = AB, P is mid-point of AB. Q is mid-point of DR and PR // BS. Prove that :
(i) AQ // BS,
(ii) DS = 3 RS.
Step 1: Since P is the mid-point of AB, AP = PB. Also, we are given 2AD = AB, which implies AD = 1/2 AB, so AD = AP = PB.
Step 2: This means the segments DA, AP, and PB on the transversal DB are all equal.
Step 3: In triangle DPR, A is the mid-point of DP and Q is the mid-point of DR.
Step 4: By the mid-point theorem, AQ is parallel to PR.
Step 5: Since it is given that PR // BS, we can conclude that AQ // BS. (Proves i).
Step 6: We now have three parallel lines AQ, PR, and BS cut by transversals DB and DS.
Step 7: Since the parallel lines make equal intercepts on transversal DB (DA = AP = PB), by intercept theorem they must make equal intercepts on transversal DS.
Step 8: Therefore, DQ = QR = RS.
Step 9: We know DS = DQ + QR + RS.
Step 10: Substituting the equals, DS = RS + RS + RS = 3 RS. (Proves ii). Hence Proved.

4. The side AC of a triangle ABC is produced to point E so that CE = 1/2 AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that :
(i) 3DF = EF
(ii) 4CR = AB.
Step 1: In triangle ABC, draw DP // AB. Since D is the mid-point of BC, by converse mid-point theorem, P is mid-point of AC. Thus, AP = PC = 1/2 AC.
Step 2: We are given CE = 1/2 AC. Therefore, PC = CE, which means C is the mid-point of segment PE.
Step 3: In triangle DPE, C is the mid-point of PE and CR is parallel to DP (since both are parallel to AB).
Step 4: By the converse of the mid-point theorem, R must be the mid-point of DE. Thus, DR = RE.
Step 5: Also by the mid-point theorem in triangle DPE, CR = 1/2 DP.
Step 6: Back in triangle ABC, by mid-point theorem, DP = 1/2 AB.
Step 7: Substituting this into step 5, CR = 1/2 × (1/2 AB) = 1/4 AB. Thus, 4CR = AB. (Proves ii).
Step 8: Now consider triangle AFE. CR is parallel to AF (AB).
Step 9: By basic proportionality, EC / EA = ER / EF.
Step 10: EA = EC + CA = CE + 2CE = 3CE. Therefore, EC / EA = 1/3.
Step 11: So ER / EF = 1/3, meaning EF = 3 ER.
Step 12: We know EF = FD + DR + RE = FD + 2RE (since DR = RE).
Step 13: Since EF = 3RE, we get FD + 2RE = 3RE, resulting in FD = RE.
Step 14: This means FD = DR = RE. So EF = FD + FD + FD = 3 DF. (Proves i). Hence Proved.

5. In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :
(i) M, A and N are collinear.
(ii) A is the mid-point of MN.
Step 1: Medians BP and CQ mean P and Q are mid-points of AC and AB respectively.
Step 2: In quadrilateral AMBC, diagonals AB and MC intersect at Q. Since CQ = QN and AQ = QB, diagonals bisect each other, making AMBN a parallelogram.
Step 3: As AMBN is a parallelogram, AM // BC and AM = BC.
Step 4: Similarly, in quadrilateral ANCB, diagonals AC and NB intersect at P. Since AP = PC and BP = PM, it's a parallelogram.
Step 5: As ANCB is a parallelogram, AN // BC and AN = BC.
Step 6: From steps 3 and 5, both AM and AN are parallel to BC and pass through a common point A.
Step 7: By parallel postulate, there can only be one line through A parallel to BC, meaning M, A, and N lie on the same straight line (collinear). (Proves i).
Step 8: Furthermore, since AM = BC and AN = BC, it follows that AM = AN.
Step 9: Since M, A, N are collinear and AM = AN, A is the mid-point of MN. (Proves ii). Hence Proved.

6. In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.
Step 1: In triangle ABC, E is the mid-point of BC and EF is given to be parallel to AB.
Step 2: By the converse of the mid-point theorem, F must be the mid-point of AC.
Step 3: Since D and F are mid-points of AB and AC respectively, by the mid-point theorem, DF is parallel to BC.
Step 4: Now consider the quadrilateral BEFD.
Step 5: We know EF is parallel to BD (since EF is parallel to AB).
Step 6: We also know DF is parallel to BE (since DF is parallel to BC).
Step 7: Since both pairs of opposite sides are parallel, quadrilateral BEFD is a parallelogram. Hence Proved.

7. In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that:
(i) triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
Step 1: ABCD is a parallelogram, so AB // DC and AB = DC.
Step 2: E and F are mid-points, so AE = EB = 1/2 AB and DF = FC = 1/2 DC.
Step 3: This makes AE // DF and AE = DF. So AEFD is a parallelogram, meaning AD // EF and AF // DE.
Step 4: Similarly, EBCF is a parallelogram, meaning EF // BC and EC // FB.
Step 5: In triangles HEB and FHC: EB = FC (halves of equal opposite sides).
Step 6: Angle EBH = Angle HFC (Alternate interior angles, since AB // DC and transversal BF).
Step 7: Angle HEB = Angle FCH (Alternate interior angles with transversal EC).
Step 8: By ASA congruence, triangle HEB is congruent to triangle FHC. (Proves i).
Step 9: Now for quadrilateral GEHF. From step 3, AF // DE, which means GF // EH.
Step 10: From step 4, EC // FB, which means GE // HF.
Step 11: Since opposite sides are parallel, GEHF is a parallelogram. (Proves ii). Hence Proved.

8. In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet side BC at points M and N respectively. Prove that : BM = MN = NC.
Step 1: Since DF // EG // BC, these parallel lines make equal intercepts on transversal AB (AD = DE = EB).
Step 2: By the Equal Intercept Theorem, they must make equal intercepts on transversal AC, so AF = FG = GC.
Step 3: Now, lines FM // GN // AB are drawn, crossing transversals AC and BC.
Step 4: Since these parallel lines make equal intercepts on transversal AC (AF = FG = GC), by the intercept theorem they must make equal intercepts on transversal BC.
Step 5: Therefore, the intercepts on BC are equal, meaning BM = MN = NC. Hence Proved.

9. In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use Intercept Theorem to show that MN bisects AD.
Step 1: Draw a line through vertex A parallel to the base BC. Let's call it line L.
Step 2: Since M and N are mid-points of AB and AC, by the mid-point theorem, MN is parallel to BC.
Step 3: We now have three parallel lines: line L, line MN, and line BC.
Step 4: Transversal AB intersects them, and since M is the mid-point, it makes equal intercepts AM = MB.
Step 5: By the intercept theorem, these parallel lines must make equal intercepts on any other transversal.
Step 6: Consider transversal AD. It intersects the three parallel lines at A, at a point on MN, and at D.
Step 7: Due to equal intercepts, the point where MN crosses AD must be the mid-point of AD.
Step 8: Therefore, MN bisects AD. Hence Proved.

10. If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle, show that the diagonals AC and BD intersect at right angle.
Step 1: Let the inner rectangle be PQRS, where P, Q, R, S are mid-points of AB, BC, CD, DA respectively.
Step 2: By the mid-point theorem, PQ // AC and QR // BD.
Step 3: Since PQRS is given to be a rectangle, the angle between adjacent sides PQ and QR is 90 degrees.
Step 4: The angle between the diagonals AC and BD is exactly the same as the angle between the lines PQ and QR because PQ // AC and QR // BD.
Step 5: Therefore, the angle between AC and BD is also 90 degrees.
Step 6: This proves that the diagonals AC and BD intersect at a right angle. Hence Proved.

TEST YOURSELF

1. Multiple Choice Type :
Choose the correct answer from the options given below.
(a) The mid-points of the sides of a triangle are joined together to get four triangles. These four triangles are :
(i) not equal to each other
(ii) congruent to each other
(iii) not congruent to each other
(iv) none of these
Step 1: By the mid-point theorem, the lines joining mid-points create an inner triangle whose sides are half the length of the main triangle's sides.
Step 2: This process divides the original triangle into four smaller triangles, all having identical side lengths (SSS congruence).
Answer: (ii) congruent to each other

(b) In the given figure, AB//CD//EF. If AC = 7 cm, AE = 14 cm and BF = 20 cm, then DF is equal to :
(i) 7 cm
(ii) 14 cm
(iii) 10 cm
(iv) 16 cm
Step 1: We are given AE = 14 cm and AC = 7 cm. This means CE = AE - AC = 14 - 7 = 7 cm.
Step 2: Since AC = CE, the parallel lines make equal intercepts on transversal AE.
Step 3: By intercept theorem, they make equal intercepts on transversal BF, so BD = DF.
Step 4: Since BF = 20 cm and BD = DF, DF = 20 / 2 = 10 cm.
Answer: (iii) 10 cm

(c) In the given figure, AB//DC//EF and E is mid-point of side AD, then :
(i) OE : OF = 1 : 3
(ii) OE = OF
(iii) OF = 2 × OE
(iv) CF = FB
Step 1: The figure is a trapezium ABCD cut by parallel line EF passing through E, the mid-point of AD.
Step 2: Since EF is parallel to the parallel sides and passes through the mid-point of one non-parallel side, it must pass through the mid-point of the other.
Step 3: Therefore, F is the mid-point of BC.
Step 4: This means CF is equal to FB.
Answer: (iv) CF = FB

(d) In rhombus PQRS; A, B and C are mid-points of sides PQ, QR and RS respectively. If angle P = 60°, the angle PQR is equal to :
(i) 60°
(ii) 90°
(iii) 120°
(iv) none of these
Step 1: In any rhombus (or parallelogram), consecutive interior angles are supplementary, meaning they add up to 180 degrees.
Step 2: Therefore, angle P + angle PQR = 180°.
Step 3: We are given angle P = 60°.
Step 4: 60° + angle PQR = 180°, so angle PQR = 120°.
Answer: (iii) 120°

(e) Statement (1) : The diagonals of a quadrilateral are perpendicular to each other; P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. Then PQRS will be a rectangle.
Statement (2) : Quadrilateral PQRS will be a square as each of its angles will be 90°.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: From a previous proof, if diagonals are perpendicular, the quadrilateral formed by mid-points is a rectangle. So, Statement 1 is true.
Step 2: A rectangle has 90-degree angles, but its adjacent sides are only equal if the original diagonals were equal in length. Therefore, it is not necessarily a square. Statement 2 is false.
Answer: (iii) Statement 1 is true, and statement 2 is false.

(f) Statement (1) : AD is median of triangle ABC and DE is parallel to BA. Then DE will bisect AC.
Statement (2) : DE is the median of triangle ADC.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: AD is a median, so D is mid-point of BC. DE is parallel to BA. By converse mid-point theorem, E is mid-point of AC. So DE bisects AC. Statement 1 is true.
Step 2: In triangle ADC, a segment from vertex D to the mid-point E of side AC is exactly the definition of a median. So Statement 2 is true.
Answer: (i) Both the statements are true.

(g) Assertion (A) : The figure formed by joining the mid-points of the sides of a quadrilateral ABCD is a square.
Reason (R) : Diagonals of the quadrilateral ABCD are not given to be equal and perpendicular to each other.
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 1: The figure formed by joining mid-points of any quadrilateral is generally a parallelogram, not a square. So Assertion A is false.
Step 2: To form a square, the original diagonals must be equal and perpendicular. Since this is not given, they are not guaranteed to be. So Reason R is a true statement.
Answer: (ii) A is false, R is true.

(h) Assertion (A) : R, S, D and E are mid-points of OC, OB, AB and AC respectively, then DERS is a parallelogram.
Reason (R) : DS // AO // ER and DS = ER = 1/2 AO.
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 1: In triangle AOB, D and S are mid-points, so DS // AO and DS = 1/2 AO.
Step 2: In triangle AOC, E and R are mid-points, so ER // AO and ER = 1/2 AO.
Step 3: Therefore, DS // ER and DS = ER, which proves DERS is a parallelogram. So both A and R are true.
Step 4: Reason R perfectly explains the parallel and equal properties needed to prove the parallelogram.
Answer: (iii) Both A and R are true and R is the correct reason for A.

2. In triangle ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm; find the perimeter of the parallelogram BDEF.
Step 1: D and E are mid-points of AB and AC. By mid-point theorem, DE is parallel to BC (so DE // BF).
Step 2: It is given that EF is drawn parallel to AB (so EF // BD).
Step 3: Since both pairs of opposite sides are parallel, BDEF is a parallelogram.
Step 4: Being a parallelogram, opposite sides are equal. BD = EF and DE = BF.
Step 5: D is mid-point of AB, so BD = 1/2 AB = 1/2 × 8 = 4 cm.
Step 6: By mid-point theorem, DE = 1/2 BC = 1/2 × 9 = 4.5 cm.
Step 7: Perimeter of BDEF = 2 × (BD + DE) = 2 × (4 + 4.5) = 2 × 8.5 = 17 cm.
Answer: 17 cm

3. P, Q and R are mid-points of sides AB, BC and CD respectively of a rhombus ABCD. Show that PQ is perpendicular to QR.
Step 1: Draw the diagonals AC and BD of the rhombus ABCD. Diagonals of a rhombus intersect at right angles.
Step 2: In triangle ABC, P and Q are mid-points of AB and BC. By mid-point theorem, PQ // AC.
Step 3: In triangle BCD, Q and R are mid-points of BC and CD. By mid-point theorem, QR // BD.
Step 4: The angle between lines PQ and QR is equal to the angle between their respective parallel diagonals AC and BD.
Step 5: Since diagonals AC and BD are perpendicular to each other (90°), PQ is perpendicular to QR. Hence Proved.

4. The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral obtained by joining the mid-points of its adjacent sides is a rectangle.
Step 1: Let P, Q, R, S be the mid-points of AB, BC, CD, DA respectively.
Step 2: By joining these mid-points, we get a parallelogram PQRS because PQ // AC // SR and PS // BD // QR.
Step 3: Since PQ // AC and QR // BD, the angle between PQ and QR is the same as the angle between AC and BD.
Step 4: It is given that the diagonals AC and BD are perpendicular (intersect at 90°).
Step 5: Therefore, the angle between adjacent sides PQ and QR of the parallelogram is 90°.
Step 6: A parallelogram with a right angle is a rectangle. Hence Proved.

5. In Δ ABC, E is mid-point of the median AD and BE produced meets side AC at point Q. Show that BE : EQ = 3 : 1.
Step 1: Draw a line through D parallel to BQ, meeting side AC at point P (DP // BQ).
Step 2: In triangle BQC, D is the mid-point of BC (because AD is a median) and DP is parallel to BQ.
Step 3: By the converse of the mid-point theorem, P is the mid-point of QC, so QP = PC.
Step 4: In triangle ADP, E is the mid-point of AD and EQ is parallel to DP (since BQ is parallel to DP).
Step 5: By the converse of the mid-point theorem, Q is the mid-point of AP, so AQ = QP.
Step 6: From steps 3 and 5, AQ = QP = PC.
Step 7: In triangle ADP, EQ = 1/2 DP (by mid-point theorem).
Step 8: In triangle BQC, DP = 1/2 BQ (by mid-point theorem). This means BQ = 2 DP.
Step 9: We know BE = BQ - EQ. Substituting, BE = 2 DP - 1/2 DP = 3/2 DP.
Step 10: Therefore, the ratio BE : EQ = (3/2 DP) : (1/2 DP) = 3 : 1. Hence Proved.

6. In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF. Show that : EF = AC.
Step 1: In triangle DEF, M and N are mid-points of DE and DF respectively.
Step 2: By the mid-point theorem, MN // EF and MN = 1/2 EF.
Step 3: In triangle ABC, M and N are mid-points of AB and BC respectively.
Step 4: By the mid-point theorem, MN // AC and MN = 1/2 AC.
Step 5: From steps 2 and 4, we have 1/2 EF = MN and 1/2 AC = MN.
Step 6: Equating the two, 1/2 EF = 1/2 AC, which simplifies to EF = AC. Hence Proved.

7. In triangle ABC: D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm. find the perimeter of the parallelogram BDEF.
Step 1: By the mid-point theorem, since D and E are mid-points of AB and AC, DE is parallel to BC (so DE // BF).
Step 2: We are given that EF is drawn parallel to AB (so EF // BD).
Step 3: Since both pairs of opposite sides are parallel, BDEF is a parallelogram.
Step 4: Opposite sides of a parallelogram are equal, so BD = EF and DE = BF.
Step 5: D is mid-point of AB, so BD = 1/2 × 16 = 8 cm.
Step 6: By mid-point theorem, DE = 1/2 × BC = 1/2 × 18 = 9 cm.
Step 7: Perimeter of BDEF = 2 × (BD + DE) = 2 × (8 + 9) = 2 × 17 = 34 cm.
Answer: 34 cm

8. In the given figure, AD and CE are medians and DF // CE. Prove that : FB = 1/4 AB.
Step 1: AD and CE are medians, so D is mid-point of BC and E is mid-point of AB.
Step 2: In triangle BCE, D is the mid-point of BC and DF is given to be parallel to CE.
Step 3: By the converse of the mid-point theorem, F must be the mid-point of BE.
Step 4: Therefore, FB = 1/2 BE.
Step 5: Since CE is a median, E is the mid-point of AB, meaning BE = 1/2 AB.
Step 6: Substituting this into step 4, FB = 1/2 × (1/2 AB) = 1/4 AB. Hence Proved.

9. In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P. Prove that :
(i) BP = 2AD
(ii) O is mid-point of AP.
Step 1: In triangle ABP, E is the mid-point of AB and EC is parallel to AP (given).
Step 2: By converse of the mid-point theorem, C is the mid-point of BP. So, BC = CP.
Step 3: We know BP = BC + CP = 2 BC.
Step 4: Since ABCD is a parallelogram, AD = BC.
Step 5: Substituting AD for BC, we get BP = 2 AD. (Proves i).
Step 6: In parallelogram ABCD, AB // DC, which means AE // OC.
Step 7: We are given AP // EC. This makes AECO a parallelogram.
Step 8: Opposite sides of a parallelogram are equal, so OC = AE. Since E is mid-point of AB, AE = 1/2 AB. Thus OC = 1/2 AB.
Step 9: We know AB = DC. So OC = 1/2 DC, meaning O is the mid-point of DC.
Step 10: In triangle ADP (wait, line from D to P?), let's use triangle CDP. No, consider transversals.
Step 11: In triangle ABP, C is mid-point of BP, and OC // AB. By converse mid-point theorem, O is mid-point of AP. (Proves ii). Hence Proved.

10. In a trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC. Prove that : AB + DC = 2EF
Step 1: Draw diagonal BD which intersects EF at point O.
Step 2: In triangle ABD, E is the mid-point of AD and EO // AB (since EF // AB).
Step 3: By the converse of the mid-point theorem, O is the mid-point of BD.
Step 4: By the mid-point theorem in triangle ABD, EO = 1/2 AB.
Step 5: In triangle BCD, O is the mid-point of BD and OF // DC (since EF // DC).
Step 6: By the converse of the mid-point theorem, F is the mid-point of BC (this aligns with the given).
Step 7: By the mid-point theorem in triangle BCD, OF = 1/2 DC.
Step 8: Adding EO and OF, we get EF = EO + OF = 1/2 AB + 1/2 DC = 1/2 (AB + DC).
Step 9: Multiplying by 2, we obtain 2EF = AB + DC. Hence Proved.

11. In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
Step 1: Since AD is a median, D is the mid-point of BC.
Step 2: In triangle ABC, D is the mid-point of BC and DE is parallel to BA.
Step 3: By the converse of the mid-point theorem, a line drawn through the mid-point of one side parallel to another side bisects the third side.
Step 4: Therefore, E must be the mid-point of AC.
Step 5: The line segment connecting a vertex (B) to the mid-point (E) of the opposite side (AC) is a median.
Step 6: Therefore, BE is also a median. Hence Proved.

12. Adjacent sides of a parallelogram are equal and one of diagonals is equal to any one of the sides of this parallelogram. Show that its diagonals are in the ratio √3 : 1.
Step 1: A parallelogram with equal adjacent sides is a rhombus. Let its sides be length 'a'. So AB=BC=CD=DA=a.
Step 2: One diagonal is equal to a side. Let diagonal BD = a.
Step 3: Diagonals of a rhombus bisect each other at right angles. Let them intersect at O.
Step 4: In right-angled triangle AOB, the hypotenuse AB = a, and side OB = BD / 2 = a / 2.
Step 5: By Pythagoras theorem, AO² + OB² = AB².
Step 6: AO² + (a/2)² = a² → AO² + a²/4 = a².
Step 7: AO² = a² - a²/4 = (3a²) / 4.
Step 8: Taking square root, AO = (a√3) / 2.
Step 9: The full diagonal AC is 2 × AO = a√3.
Step 10: The ratio of the diagonals AC : BD is (a√3) : a = √3 : 1. Hence Proved.

Case-Study Based Question
A school is designing a triangular garden ΔABC To construct a walking path inside the garden, the gardener marks the mid-points of two sides: point D is the mid-point of side AB and point E is the mid-point of side AC The path DE is drawn to connect these mid-points. The length of side BC of the triangular garden is 12 m, AB = 10 m and AC = 10 m.
Based on the above information answer the following:
(i) What is the length of path DE?
(ii) Assign a special name to Quadrilateral BCED and find its perimeter.
Step 1: For (i): D and E are mid-points of AB and AC. By the mid-point theorem, DE is parallel to BC and is half its length.
Step 2: DE = 1/2 × BC = 1/2 × 12 = 6 m. Length of DE is 6 m.
Step 3: For (ii): Since DE is parallel to BC, the quadrilateral BCED has one pair of parallel opposite sides, making it a trapezium.
Step 4: D is the mid-point of AB (10 m), so BD = 5 m. E is the mid-point of AC (10 m), so CE = 5 m.
Step 5: Since the non-parallel sides are equal (BD = CE = 5 m), it is specifically an Isosceles Trapezium.
Step 6: Perimeter of BCED = BC + CE + ED + DB = 12 + 5 + 6 + 5 = 28 m.
Answer: (i) 6 m (ii) Isosceles Trapezium, 28 m