**Chapter 5 - Quadrilaterals**

**Practice Set 5.1**

**Practice Set 5.1**

**1. ****Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠XYZ = 135⁰ then what is the measure of ∠XWZ and ∠YZW ? If l(OY) = 5 cm then l(WY) = ?**

**Given:**

**□ WXYZ is a parallelogram**

**Diagonal WY and Diagonal XZ intersect at Point O**

**∠XYZ = 135⁰**

**l(OY) = 5 cm**

** **

**To find:**

**∠XWZ, ∠YZW and l(WY)**

** **

**Solution:**

**□ WXYZ is a parallelogram [Given]**

**∴ ∠XYZ ≅ ∠XWZ [Opposite angles of a parallelogram are congruent]**

**∵ ∠XYZ = 135⁰**

**∴ ∠XWZ = 135⁰ ……(i)**

** **

**∠YZW + ∠XYZ = 180⁰ [Adjacent angles of a parallelogram are supplementary]**

**∴ ∠YZW + 135= 180⁰ [From (i)]**

**∴ ∠YZW = 180⁰ – 135⁰**

**∴ ∠YZW = 45⁰**

** **

**Now,**

**l(OY) ≅ l(WY) [Diagonals of a parallelogram bisect each other]**

**∵ l(OY) = 5 cm [Given]**

**∴ l(WY) = 5 cm**

** **

**l(OY) + l(WY) = l(OY) [From Figure]**

**5 + 5 = l(OY)**

**∴ l(OY) = 10 cm**

Ans: ∠XWZ = 135⁰, ∠YZW = 45⁰, l(WY) = 10 cm

**2. In a parallelogram ABCD, If ∠A = (3x + 12)⁰, ∠B = (2x – 32)⁰ then find the value of x and then find the measures of ∠C and ∠D. **

**Given:**

**□ABCD is a parallelogram **

**∠A = (3x + 12)⁰**

**∠B = (2x – 32)⁰**

** **

**To find:**

**x, ∠C and ∠D**

** **

**Solution:**

**□ABCD is a parallelogram [Given]**

**∴ ∠A + ∠B = 180⁰ [Adjacent angles of a parallelogram are supplementary]**

**∴ (3x + 12)⁰ + (2x – 32)⁰ = 180⁰**

**∴ 3x + 12 + 2x – 32 = 180**

**∴ 5x – 20 = 180**

**∴ 5x= 180 + 20**

**∴ 5x = 200**

**∴ x = \( \frac{200}{5} \) **

**∴ x = 40**

** **

**∴ The value of ∠A and ∠B is**

**∠A = (3x + 12)⁰**

**∠A = [3(40) + 12]⁰**

**∴ ∠A = (120 + 12)⁰**

**∴ ∠A = 132⁰**

** **

**∠B = (2x – 32)⁰**

**∴ ∠B = [2(40) – 32]⁰**

**∴ ∠B = (80 – 32)⁰**

**∴ ∠B = 48⁰**

** **

**∴ ∠C ≅ ∠A [Opposite angles of a parallelogram]**

**∵ ∠C = 135⁰**

**∴ ∠A = 135⁰**

** **

**∴ ∠B ≅ ∠D [Opposite angles of a parallelogram]**

**∵ ∠B = 48⁰**

**∴ ∠D = 48⁰**

**Ans: **The value of x is 40, and the measures of ∠C and ∠D are 132⁰ and 48⁰ respectively.

**3. Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides. **

**Given:**

**Perimeter of □ABCD = 150 cm**

**l(AB) = l(AD) + 25 cm**

** **

**To find:**

**l(AB), l(BC), l(CD) and l(AD)**

** **

**Solution:**

**Let l(AD) = x**

**∴ l(AB) = l(AD) + 25 cm**

**∴ l(AB) = x + 25 cm**

**∴ l(AB) ≅ l(CD) [Opposite sides of a parallelogram are congruent]**

**∵ l(AB) = x + 25 cm**

**∴ l(CD) = x + 25 cm**

** **

**∴ l(AD) ≅ l(BC) [Opposite sides of a parallelogram are congruent]**

**∵ l(AD) = x**

**∴ l(BC) = x**

** **

**Now,**

**Perimeter of □ABCD = 150 cm [Given]**

**∴ l(AB + l(BC) + l(DC) + l(AD) = 150**

**∴ (x + 25) + x + (x + 25) + x = 150**

**∴ 4x + 50 = 150**

**∴ 4x = 150 – 50**

**∴ 4x = 100**

**∴ x = \( \frac{100}{4} \) **

**∴ x = 25**

** **

**Now, the values of **

**l(AD) = l(BC) = x = 25 cm**

**l(AB) = l(DC) = x + 25 = 25 + 25 = 50 cm**

**Ans: **The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.

**4. If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram. **

**Given:**

**Adjacent angles of a parallelogram is 1 : 2**

** **

**To find:**

**∠A, ∠B, ∠C and ∠D**

** **

**Solution:**

**Let the common multiple be x.**

**∴ ∠A = x⁰ and ∠B = 2x⁰ [Given condition]**

** **

**□ABCD is a parallelogram [Given]**

**∴ ∠A + ∠B = 180⁰ [Adjacent angles of a parallelogram are supplementary]**

**∴ x + 2x = 180**

**∴ 3x = 180**

**∴ x = \( \frac{180}{3} \) **

**∴ x = 60**

** **

**∴ The value of**

**∠A = x⁰ = 60⁰**

**∠B = 2x⁰ = 2 x 60⁰ = 120⁰**

** **

**∴ ∠A ≅ ∠C [Opposite angles of a parallelogram]**

**∵ ∠A = 60⁰**

**∴ ∠C = 60⁰**

** **

**∴ ∠B ≅ ∠D [Opposite angles of a parallelogram]**

**∵ ∠B = 120⁰**

**∴ ∠D = 120⁰**

**Ans:** The measures of the angles of the parallelogram are 60⁰, 120⁰, 60⁰ and 120⁰.

**5*. Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that □ ABCD is a rhombus. **

**Given: **

**l(AO) = 5**

**l(BO) = 12**

**l(AB) = 13**

** **

**To prove: **

**□ABCD is a rhombus.**

** **

**Proof:**

**l(AO) = 5 [Given]**

**l(BO) = 12 [Given]**

**l(AB) = 13 [Given]**

** **

**Now,**

**AO² + BO² = 5² + 12²**

**AO² + BO² = 25 + 144**

**∴ AO² + BO² = 169 ……(i)**

** **

**And,**

**AB² = 13² = 169 ……(ii)**

** **

**From (i) and (ii)**

**∴ AB² = AO² + BO² **

**∴ ∆ AOB is a right-angled triangle [By Converse of Pythagoras theorem]**

**∴ ∠AOB = 90⁰**

**∴ seg AC ⊥ seg BD …..(iii) [A – O – C]**

**∴ □ABCD is a rhombus [The diagonals of a rhombus always bisect each other at 90⁰]**

** **

**Hence Proved.**

**6. In the figure 5.12, □PQRS and □ABCR are two parallelograms. If ∠P = 110⁰ then find the measures of all angles of □ABCR. **

**Given: **

**∠P = 110⁰**

** **

**To find: **

**∠A, ∠B, ∠C and ∠R**

** **

**Solution:**

**□PQRS is a parallelogram [Given]**

**∴ ∠P ≅ ∠R [Opposite angles of a parallelogram]**

**∵ ∠P = 110⁰**

**∴ ∠R = 110⁰ …..(i)**

** **

**□ABCR is a parallelogram [Given]**

**∴ ∠A + ∠R = 180⁰ [Adjacent angles of a parallelogram are supplementary]**

**∴ ∠A + 110⁰ = 180⁰ [From (i)]**

**∴ ∠A= 180⁰ – 110⁰**

**∴ ∠A = 70⁰**

** **

**∠C ≅ ∠A [Opposite angles of a parallelogram]**

**∵ ∠C = 70⁰**

**∴ ∠A = 70⁰**

** **

**∠R ≅ ∠B [Opposite angles of a parallelogram]**

**∵ ∠R = 110⁰**

**∴ ∠B = 110⁰**

**Ans: **∠A = 70⁰, ∠B = 110⁰, ∠C = 70⁰ and ∠R = 110⁰

**7. In figure 5.13 □ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.**

**Given: **

**□ABCD is a parallelogram**

**BE = AB**

** **

**To prove: **

**FC = FB**

** **

**Proof:**

**□ABCD is a parallelogram [Given]**

**∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram]**

**seg AB ≅ seg BE ……..(ii) [Given]**

**seg DC ≅ seg BE ……..(iii) [From (i) & (ii)]**

** **

**side DC || side AB [Opposite sides of a parallelogram]**

**i.e. side DC || seg AE and seg DE is their transversal [A – B – E]**

**∴ ∠CDE ≅ ∠AED [Alternate Angles]**

**∴ ∠CDF ≅ ∠BEF …..(iv) [D – F – E, A – B – E]**

** **

**In ∆DFC and ∆EFB,**

**seg DC ≅ seg EB [From (iii)]**

**∠CDF ≅ ∠BEF [From (iv)]**

**∠DFC ≅ ∠EFB [Vertically opposite angles]**

**∴ ∆DFC ≅ ∆EFB [SAA test]**

**∴ FC ≅ FB [c.s.c.t]**

** **

**∴ Line ED bisects seg BC at point F**

** **

**Hence Proved.**

**Practice Set 5.2**

**Practice Set 5.2****1. In figure 5.22, □ABCD is a parallelo****gram, P and Q are midpoints of side AB ****and DC respectively, then prove □APCQ ****is a parallelogram. **

**Given: **

**□ABCD is a parallelogram**

**P and Q are midpoints of side AB **

**and DC**

** **

**To prove: **

**□APCQ is a parallelogram**

** **

**Proof:**

**AB ≅ AP [P is the midpoint of side AB]**

**∴ AP = \( \frac{1}{2} \) AB …..(i) **

** **

**QC ≅ DQ [Q is the midpoint of side DC]**

**∴ QC = \( \frac{1}{2} \) DC …..(ii) **

** **

**□ABCD is a parallelogram [Given]**

**∴ AB ≅ DC [Opposite sides of a parallelogram]**

**∴ \( \frac{1}{2} \) AB = \( \frac{1}{2} \) DC [Multiplying both sides by \( \frac{1}{2} \)]**

**∴ AP = QC ….(iii) [From (i) and (ii)]**

** **

**Also, **

**Side AB || Side DC [Opposite sides of a parallelogram]**

**i.e. AP || QC ….(iv) [A – P – B, D – Q – C]**

** **

**We know that,**

**A quadrilateral is a parallelogram if its opposite sides is parallel and congruent**

** **

**∴ □APCQ is a parallelogram. [From (iii) & (iv)]**

** **

**Hence Proved.**

**2. Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram. **

**Given:**

**□ABCD is a rectangle**

** **

**To prove: **

**Rectangle ABCD is a parallelogram**

** **

**Proof:**

**□ABCD is a rectangle**

**∴ ∠A ≅ ∠B ≅ ∠C ≅ ∠D = 90° [Angles of a rectangle]**

** **

**We know that, **

**A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent**

**∴ Rectangle ABCD is a parallelogram.**

** **

**Hence Proved.**

**3. In figure 5.23, G is the point of concurrence of medians of D DEF. Take point H on ray DG such that D-G-H and ****DG = GH, then prove that □GEHF is a parallelogram. **

**Given: **

**Point G (centroid) is the point of concurrence of the medians of △DEF**

**DG = GH**

** **

**To prove: **

**□GEHF is a parallelogram.**

** **

**Proof:**

**Let ray DH intersect seg EF at point A such that E – A – F**

**∴ seg DA is the median of ∆DEF**

**∴ EA ≅ FA ……(i)**

** **

**Point G is the centroid of ∆DEF**

**∴ DG:GI = 2:1 [Centroid divides each median in the ratio 2:1]**

**∴ DG = 2(GI)**

**∴ GH = 2(GI) [DG ≅ GH, Given]**

**∴ GA + HA = 2(GA) [G – A – H]**

**∴ HA = 2(GA) – GA**

**∴ HA = GA ….(ii)**

** **

**We know that,**

**A quadrilateral is a parallelogram, if its diagonals bisect each other**

**∴ □GEHF is a parallelogram [From (i) & (ii)]**

** **

**Hence Proved.**

**4. Prove that a quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle. **

**Given: **

**□ABCD is a parallelogram.**

**Ray AS bisect ∠A**

**Ray BQ bisect ∠B**

**Ray CQ bisect ∠C**

**Ray DS bisect ∠D**

** **

**To prove: **

**□PQRS is a rectangle.**

** **

**Proof:**

**∠BAS ≅ ∠DAS [Ray AS bisects ∠A] **

**Let, ∠BAS = x°**

**∴ ∠DAS = x° ..…(i) **

** **

**∠ABQ ≅ ∠CBQ [Ray BQ bisects ∠B]**

**Let, ∠ABQ = y°**

**∴ ∠CBQ = y° ..…(ii) **

** **

**∠BCQ ≅ ∠DCQ [Ray CQ bisects ∠C]**

**Let, ∠BCQ = u°**

**∴ ∠DCQ = u° ..…(iii) **

** **

**∠ADS ≅ ∠CDS [Ray DS bisects ∠D]**

**Let, ∠ADS = v°**

**∴ ∠CDS = v° ..…(iv) **

** **

**□ABCD is a parallelogram [Given]**

**∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]**

**∴ ∠BAS + ∠DAS + ∠ABQ + ∠CBQ = 180° [Angle addition property]**

**∴ x° + x° + y° + y° = 180 [From (i) and (ii)]**

**∴ 2x° + 2y° = 180**

**∴ 2(x° + y°) = 180**

**∴ (x° + y°) = \( \frac{180}{2} \)**

**∴ x° + y° = 90° ……(v) **

** **

**Also, **

**∠C + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]**

**∴ ∠ABQ + ∠CBQ + ∠BCQ + ∠DCQ = 180° [Angle addition property]**

**∴ y° + y° + u° + u° = 180° [From (ii) and (iii)]**

**∴ 2y° + 2u° = 180°**

**∴ 2(y° + u°) = 180**

**∴ (y° + u°) = \( \frac{180}{2} \)**

**∴ y° + u° = 90° ……(vi) **

** **

**Also,**

**∠C + ∠D = 180° [Adjacent angles of a parallelogram are supplementary]**

**∴ ∠BCQ + ∠DCQ + ∠ADS + ∠CDS = 180° [Angle addition property]**

**∴ u° + u° + v° + v° = 180° [From (iii) and (iv)]**

**∴ 2u° + 2v° = 180°**

**∴ 2(u° + v°) = 180**

**∴ (u° + v°) = \( \frac{180}{2} \)**

**∴ u° + v° = 90° ……(vii) **

** **

**Also,**

**∠A + ∠D = 180° [Adjacent angles of a parallelogram are supplementary]**

**∴ ∠BAS + ∠DAS + ∠ADS + ∠CDS = 180° [Angle addition property]**

**∴ x° + x° + v° + v° = 180° [From (i) and (iv)]**

**∴ 2x° + 2v° = 180°**

**∴ 2(x° + v°) = 180**

**∴ (x° + v°) = \( \frac{180}{2} \)**

**∴ x° + v° = 90° ……(viii) **

** **

**In ∆ARB,**

**∠RAB + ∠RBA + ∠ARB = 180° [Sum of the measures of the angles of a triangle is 180°]**

**∴ x° + y° + ∠SRQ = 180° [A – S – R, B – Q – R]**

**∴ 90° + ∠SRQ = 180° [From (v)]**

**∴ ∠SRQ = 180°- 90°**

**∴ ∠SRQ = 90° …..(ix)**

** **

**In ∆DPC,**

**∠PDC + ∠PCD + ∠DPC = 180° [Sum of the measures of the angles of a triangle is 180°]**

**∴ v° + u° + ∠SPQ = 180° [P – Q – C, P – S – D]**

**∴ 90° + ∠SPQ = 180° [From (vii)]**

**∴ ∠SPQ = 180°- 90°**

**∴ ∠SPQ = 90° …..(x)**

** **

**In ∆ASD,**

**∠ASD + ∠SAD + ∠SDA = 180° [Sum of the measures of angles a triangle is 180°]**

**∴ ∠ASD + x° + v° = 180° **

**∴ ∠ASD + 90° = 180° [From (viii)]**

**∴ ∠ASD = 180°- 90°**

**∴ ∠ASD = 90°**

** **

**∴ ∠PSR ≅ ∠ASD [Vertically opposite angles]**

**∵ ∠ASD = 90°**

**∴ ∠PSR = 90° …..(xi)**

** **

**In ∆BQC,**

**∠BQC + ∠CBQ + ∠BCQ = 180° [Sum of the measures of angles a triangle is 180°]**

**∴ ∠BQC + y° + u° = 180° [From (vi)]**

**∴ ∠BQC + 90° = 180°**

**∴ ∠BQC = 180°- 90°**

**∴ ∠BQC = 90°**

** **

**∴ ∠BQC ≅ ∠PQR [Vertically opposite angles]**

**∵ ∠BQC = 90°**

**∴ ∠PQR = 90° …..(xii)**

** **

**∴ In □PQRS,**

**∠SRQ ≅ ∠SPQ ≅ ∠PSR ≅ ∠PQR = 90° [From (ix), (x), (xi), (xii)]**

**∴ □PQRS is a rectangle [Each angle is a right angle]**

** **

**Hence Proved.**

**5. In figure 5.25, if points P, Q, R, S are ****on the sides of parallelogram such that ****AP = BQ = CR = DS then prove that ****□PQRS is a parallelogram.**

**Given: **

**□ABCD is a parallelogram**

**AP = BQ = CR = DS**

** **

**To prove: **

**□PQRS is a parallelogram**

** **

**Proof:**

**□ABCD is a parallelogram [Given]**

**∴ ∠B ≅ ∠D ….(i) [Opposite angles of a parallelogram are congruent]**

** **

**∴ ∠A ≅ ∠C ….(ii) [Opposite angles of a parallelogram are congruent]**

** **

**Also, **

**AB ≅ CD [Opposite sides of a parallelogram are congruent]**

**∴ AP + BP = DR + CR [A – P – B, D – R – C]**

**∴ AP + BP = DR + AP [Given, AP = CR]**

**∴ BP = DR ….(iii)**

** **

**AD ≅ BC [Opposite sides of a parallelogram are congruent]**

**∴ AS + SD = BQ + QC [A – S – D, B – Q – C]**

**∴ AS + BQ = BQ + QC [Given, BQ = DS]**

**∴ AS = QC ….(iv)**

** **

**In ∆PBQ and ∆RDS**

**seg BP ≅ seg DR [From (iii)]**

**∠PBQ ≅ ∠RDS [From (i)]**

**seg BQ ≅ seg DS [Given]**

**∴ ∆PBQ ≅ ∆RDS [SAS test]**

**∴ seg PQ ≅ seg RS …..(v) [c.s.c.t]**

** **

**In ∆PAS ≅ ∆RCQ,**

**seg AS ≅ seg QC [From (iv)]**

**∠PAS ≅ ∠RCQ [From (ii)]**

**seg AP ≅ seg CR [Given]**

**∴ ∆PAS ≅ ∆RCQ [SAS test]**

**∴ seg PS ≅ seg RQ ….(vi) [c.s.c.t]**

** **

**We know that,**

**A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent**

** **

**∴ □PQRS is a parallelogram [From (v) & (vi)]**

** **

**Hence Proved.**

**Practice Set 5.3**

**Practice Set 5.3****1. Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find the length of BO and if ∠CAD = 35⁰ then find the measure of ∠ACB.**

**Given:**

**□ABCD is a rectangle **

**AC = 8**

**∠CAD = 35⁰**

** **

**To find: **

**l(BO) and ∠ACB**

** **

**Solution:**

**□ABCD is a rectangle [Given]**

**∴ BD = AC [Diagonals of a rectangle are congruent]**

**∵ AC = 8 cm [Given]**

**∴ BD = 8 cm [From (i)]**

** **

**BO = \( \frac{1}{2} \) BD [Diagonals of a rectangle bisect each other]**

**∴ BO = \( \frac{1}{2} \) x 8**

**∴ BO = \( \frac{8}{2} \)**

**∴ BO = 4 cm**

** **

**Now, **

**side AD || side BC and seg AC is their transversal [Opposite sides of a rectangle are parallel]**

**∴ ∠ACB ≅ ∠CAD [Alternate angles]**

**∵ ∠CAD = 35°[Given]**

**∴ ∠ACB = 35° **

** **

**Ans: **BO = 4 cm, ∠ACB = 35°

**2. In a rhombus PQRS if PQ = 7.5 then find the length of QR. If ∠QPS = 75⁰ then find the measure of ∠PQR and ∠SRQ. **

**Given:**

**□PQRS is a rhombus **

**PQ = 7.5**

**∠QPS = 75⁰**

** **

**To find: **

**l(QR), ∠PQR and ∠SRQ**

** **

**Solution:**

**□PQRS is a rhombus [Given]**

**∴ QR ≅ PQ [All sides of a rhombus are congruent]**

**∵ PQ = 7.5 cm**

**∴ QR = 7.5 cm**

** **

**Now,**

**∠QPS + ∠PQR = 180° [Adjacent angles of a rhombus are supplementary]**

**∴ 75° + ∠PQR = 180°**

**∴ ∠PQR = 180° – 75°**

**∴ ∠PQR = 105°**

**Also,**

**∠SRQ ≅ ∠QPS [Opposite angles of a rhombus are congruent]**

**∵ ∠QPS = 75⁰**

**∴ ∠SRQ = 75°**

** **

**Ans: **QR = 7.5 cm, ∠PQR = 105°& ∠SRQ = 75°

**3. Diagonals of a square IJKL intersects at point M, Find the measures of ∠IMJ, ∠JIK and ∠LJK .**

**Given:**

**□IJKL is a square **

** **

**To find: **

**∠IMJ, ∠JIK and ∠LJK**

** **

**Solution:**

**□IJKL is a square [Given]**

**∴ seg IK ⊥ seg JL [Diagonals of a square are perpendicular to each other]**

**∴ ∠ IMJ = 90°**

**and ∠ JIL 90° ……. (i) [Angle of a square]**

** **

**∠JIK = \( \frac{1}{2} \) ∠JIL [Diagonals of a square bisect the opposite angles]**

**∠JIK = \( \frac{1}{2} \) (90°) [From (i)]**

**∴ ∠JIK = 45°**

** **

**∠IJK = 90° (ii) [Angle of a square]**

** **

**iii. ∠LJK = \( \frac{1}{2} \) ∠IJK [Diagonals of a square bisect the opposite angles]**

**∴ ∠LJK = \( \frac{1}{2} \) (90°) [From (ii)]**

**∴ ∠LJK = 45°**

** **

**Ans:** ∠LJK = 90°, ∠JIK = 45°& ∠LJK = 45°

**4. Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of the rhombus and its perimeter**

**Given:**

**Let □ABCD be the rhombus **

**∴ AC = 20 cm, BD = 21 cm**

** **

**To find: **

**Side of the rhombus and its Perimeter**

** **

**Solution:**

**□ABCD be the rhombus**

**∴ AO = \( \frac{1}{2} \) AC [Diagonals of a rhombus bisect each other]**

**∴ AO = \( \frac{1}{2} \) 20**

**∴ AO = \( \frac{20}{2} \)**

**∴ AO = 10 cm**

** **

**And,**

**∴ BO = \( \frac{1}{2} \) BD [Diagonals of a rhombus bisect each other]**

**∴ BO = \( \frac{1}{2} \) 21**

**∴ BO = \( \frac{21}{2} \) cm**

** **

**In ∆AOB, **

**∠AOB = 90° [Diagonals of a rhombus are perpendicular to each other]**

**∴ ∆AOB is a right angled triangle**

** **

**∴ AB² = AO² + BO² [By Pythagoras theorem]**

**∴ AB² = (10)² + (\( \frac{21}{2} \))²**

**∴ AB² = 100 + (\( \frac{441}{4} \))**

**∴ AB² = (\( \frac{440}{4} \)) + (\(\frac{441}{4} \)) [Equalizing the denominator]**

**∴ AB² = (\( \frac{440\; +\; 441}{4} \)) **

**∴ AB² = (\( \frac{841}{4} \)) **

**∴ √(AB²) = \( \sqrt{\frac{841}4} \) [Taking square root on both sides]**

**∴ AB = \( \sqrt{\frac{841}4} \)**

**∴ AB = (\( \frac{29}{2} \)) **

**∴ AB = 14.5**

** **

**Now,**

**Perimeter of □ABCD = 4 x AB **

**Perimeter of □ABCD = 4 x 14.5 **

**Perimeter of □ABCD = 58 cm**

**Ans:** The side and perimeter of the rhombus are 14.5 cm and 58 cm respectively.

**5. State with reasons whether the following statements are ‘true’ or ‘false’.**

**(i) Every parallelogram is a rhombus.**

**Ans: **False

**All the sides of a rhombus are congruent, while only the opposite sides of a parallelogram are congruent, hence every parallelogram cannot be a rhombus.**

** **

**(ii) Every rhombus is a rectangle.**

**Ans:** False

**All the angles of a rectangle are congruent, while only opposite angles of a rhombus are congruent, hence every rhombus cannot be a rectangle.**

** **

**(iii) Every rectangle is a parallelogram. **

**Ans: **True

**The opposite sides of a parallelogram are parallel and congruent. Also, its opposite angles are congruent. The opposite sides of a rectangle are parallel and congruent. Also, all its angles are congruent. Hence, every rectangle is also a parallelogram. **

** **

**(iv) Every square is a rectangle.**

**Ans:** True

**The opposite sides of a rectangle are parallel and congruent. Also, all its angles are congruent. All the sides of a square are parallel and congruent. Also, all its angles are congruent. All the sides of a square are parallel and congruent. Also, all its angles are congruent. Hence, every square is also a rectangle.**

** **

**(v) Every square is a rhombus. **

**Ans:** True

**All the sides of a rhombus are congruent. Also, its diagonals are perpendicular bisectors of each other. All the sides of a square are congruent. Also, its diagonals are perpendicular bisectors of each other. Hence, every square is also a rhombus. **

** **

**(vi) Every parallelogram is a rectangle.**

**Ans:** False

**All the angles of a rectangle are congruent, while only the opposite angles of a parallelogram are congruent. Hence every parallelogram is not a rectangle.**

**Practice Set 5.4**

**Practice Set 5.4****1. In □IJKL, side IJ || side KL ∠I = 108⁰ ∠K = 53⁰ then find the measures of ∠J and ∠L. **

**Given:**

**side IJ || side KL **

**∠I = 108⁰**

**∠K = 53⁰**

** **

**To find: **

** ∠J and ∠L**

** **

**Solution:**

**In □IJKL,**

**side IJ || side KL and side IL is their transversal [Given]**

**∴ ∠I + ∠L = 180° [Interior angles]**

**∴ 108° + ∠L = 180°**

**∴ ∠L = 180° – 108°**

**∴ ∠L = 72°**

** **

**Also,**

**side IJ || side KL and side JK is their transversal [Given]**

**∴ ∠J + ∠K = 180° [Interior angles]**

**∴ ∠J + 53° = 180°**

**∴ ∠J = 180°- 53°**

**∴ ∠J = 127°**

** **

**Ans:** ∠L = 72°, ∠J = 127°

**2. In □ABCD, side BC || side AD, side AB ≅ side DC. If ∠A = 72⁰ then find the measures of ∠B, and ∠D. **

**Given:**

**side BC || side AD**

**side AB ≅ side DC**

**∠A = 72⁰**

** **

**To find: **

** ∠B and ∠D**

** **

**Solution:**

**In □ABCD, **

**side BC || side AD and side AB is their transversal [Given]**

**∴ ∠A + ∠B = 180° [Interior angles]**

**∴ 72° +∠B = 180°**

**∴ ∠B = 180° – 72° **

**∴ ∠B = 108°**

** **

**In ∆BPA and ∆CQD,**

**∠BPA ≅ ∠CQD [Both are right angles]**

**Hypotenuse AB ≅ Hypotenuse DC [Given]**

**seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]**

**∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]**

**∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]**

**i.e. ∠A = ∠D**

**∵ ∠A = 72⁰ [Given]**

**∴ ∠D = 72°**

**Ans:** ∠B = 108°, ∠D = 72°

**3. In □ABCD, side BC < side AD, side BC || side AD and if side BA ≅ side CD ****then prove that ∠ABC ≅ ∠DCB.**

**Given: **

**side BC < side AD**

**side BC || side AD**

**side BA = side CD**

** **

**To prove: **

**∠ABC ≅ ∠DCB**

** **

**Construction: **

**Draw seg BP ⊥ side AD, A – P – D**

**seg CQ ⊥ side AD, A – Q – D**

** **

**Proof:**

**In ∆BPA and ∆CQD,**

**∠BPA ≅ ∠CQB [Both are right angles]**

**Hypotenuse BA ≅ Hypotenuse CD [Given]**

**seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]**

**∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]**

**∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]**

**i.e. ∠A = ∠D ….(i)**

** **

**Now, **

**side BC || side AD and side AB is their transversal [Given]**

**∴ ∠A + ∠B = 180°…..(ii) [Interior angles]**

** **

**Also, **

**side BC || side AD and side CD is their transversal [Given]**

**∴ ∠C + ∠D = 180° …..(iii) [Interior angles]**

** **

**∴ ∠A + ∠B = ∠C + ∠D [From (ii) and (iii)]**

**∴ ∠A + ∠B = ∠C + ∠A [From (i)]**

**∴ ∠B = ∠C**

**i.e. ∠ABC ≅ ∠DCB**

** **

**Hence proved. **

**Practice Set 5.5**

**Practice Set 5.5****1. In figure 5.38, points X, Y, Z are the midpoints of side AB, side BC and side AC of ∆ABC respectively. AB = 5 cm, AC = 9 cm and BC = 11 cm. Find the length of XY, YZ, XZ. **

**Given: **

**X is the midpoint of side AB**

**Y is the midpoint of side BC**

**Z is the midpoint of side AC**

**AB = 5 cm**

**AC = 9 cm**

**BC = 11 cm**

** **

**To find: **

**l(XY), l(YZ) and l(XZ)**

** **

**Solution: **

**Points X and Y are the midpoints of sides AB and BC respectively [Given]**

**∴ XY = \( \frac{1}{2} \) AC [Midpoint theorem]**

**∴ XY = \( \frac{1}{2} \) x 9 [Given]**

**∴ XY = \( \frac{9}{2} \)**

**∴ XY = 4.5 cm**

** **

**Points Y and Z are the midpoints of sides BC and AC respectively [Given]**

**∴ YZ = \( \frac{1}{2} \) AB [Midpoint theorem]**

**∴ YZ = \( \frac{1}{2} \) x 5 [Given]**

**∴ YZ = \( \frac{5}{2} \)**

**∴ YZ = 2.5 cm**

** **

**Points X and Z are the midpoints of sides AB and AC respectively [Given]**

**∴ XZ = \( \frac{1}{2} \) BC [Midpoint theorem]**

**∴ XZ = \( \frac{1}{2} \) x 11 [Given]**

**∴ XZ = \( \frac{11}{2} \)**

**∴ XZ = 5.5 cm**

** **

**Ans:** l(XY) = 4.5 cm, l(YZ) = 2.5 cm & l(XZ) = 5.5 cm

**2. In figure 5.39, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR then prove that, **

**(i) SL = LR, **

**(ii) LN = (\( \frac{1}{2} \)) SQ**

**Given: **

**□PQRS and □MNRL are rectangles**

**M is the midpoint of side PR**

** **

**To prove:**

**SL = LR****LN = (\( \frac{1}{2} \)) SQ**

** **

**Proof:**

**□PQRS and □MNRL are rectangles [Given]**

**∴ ∠S ≅ ∠L ≅ ∠Q ≅ ∠N = 90° [Angles of rectangle]**

**∴ seg ML || seg PS …(i) [Corresponding angles test]**

**∴ seg MN || seg PQ …(ii) [Corresponding angles test]**

** **

**In ∆PRS,**

**Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]**

**∴ Point L is the midpoint of seg SR ……(ii) [Converse of midpoint theorem]**

**∴ SL = LR**

** **

**In ∆PRQ,**

**Point M is the midpoint of PR and seg MN || seg PQ [Given] [From (ii)]**

**∴ Point N is the midpoint of seg RQ ……(iii) [Converse of midpoint theorem]**

**∴ RN = NQ**

** **

**In ∆RSQ,**

**Points L and N are the midpoints of seg SR and seg QR respectively [From (ii) and (iii)]**

**∴ LN = \( \frac{1}{2} \) SQ [Midpoint theorem]**

** **

**Hence proved. **

**3. In figure 5.40, ∆ABC is an equilateral triangle. Points F,D and E are midpoints of side AB, side BC, side AC respectively. Show that ∆FED is an equilateral triangle. **

**Given: **

**∆ABC is an equilateral triangle**

**Point F is the midpoint of side AB**

**Point D is the midpoint of side BC**

**Point E is the midpoint of side AC**

** **

**To prove: **

**∆FED is an equilateral triangle.**

** **

**Proof:**

**∆ABC is an equilateral triangle [Given]**

**∴ AB ≅ BC ≅ AC ….(i) [Sides of an equilateral triangle]**

** **

**Point F is the midpoint of side AB and Point D is the midpoint of side BC [Given]**

**∴ FD = \( \frac{1}{2} \) AC [Midpoint theorem] …..(ii)**

** **

**Point D is the midpoint of side BC and Point E is the midpoint of side AC [Given]**

**∴ DE = \( \frac{1}{2} \) AB [Midpoint theorem]**

**i.e. DE = \( \frac{1}{2} \) AC [From (i)] …..(iii)**

** **

**Point F is the midpoint of side AB and Point E is the midpoint of side AC [Given]**

**∴ FE = \( \frac{1}{2} \) BC**

**i.e. FE = \( \frac{1}{2} \) AC [From (i)] …..(iv)**

** **

**∴ FD = DE = FE [From (ii), (iii) and (iv)]**

**∴ ∆FED is an equilateral triangle**

** **

**Hence proved.**

**4. In figure 5.41, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \( \frac{PM}{PR} \) = \( \frac{1}{3} \)**

**[Hint : draw DN || QM]**

**Given: **

**seg PD is a median of ∆PQR**

**Point T is the midpoint of seg PD.**

** **

**To Prove: **

**\( \frac{PM}{PR} \) = \( \frac{1}{3} \)**

** **

**Construction: **

**Draw seg DN || seg QM such that P – M – N and M – N – R**

** **

**Proof:**

**In ∆PDN,**

**Point T is the midpoint of seg PD and seg TM || seg DN [Given]**

**∴ Point M is the midpoint of seg PN [Construction and Q – T – M]**

**∴ PM ≅ MN [Converse of midpoint theorem]**

** **

**In ∆QMR,**

**Point D is the midpoint of seg QR and seg DN || seg QM [Construction]**

**∴ Point N is the midpoint of seg MR [Converse of midpoint theorem]**

**∴ RN ≅ MN …..(ii)**

** **

**∴ PM ≅ MN ≅ RN …..(iii) [From (i) and (ii)]**

** **

**Now, **

**PR = PM + MN + RN **

**∴ PR = PM + PM + PM [From (iii) ]**

**∴ PR = 3PM**

**∴ \( \frac{PM}{PR} \) = \( \frac{1}{3} \)**

** **

**Hence Proved. **

**Problem Set 5**

**Problem Set 5****1. Choose the correct alternative answer and fill in the blanks.**

**1. Choose the correct alternative answer and fill in the blanks.****(i) If all pairs of adjacent sides of a quadrilateral are congruent then it is called ….**

**(A) rectangle **

**(B) parallelogram **

**(C) trapezium**

**(D) rhombus**

**Answer:**

**OPTION (D)** – rhombus

** **

**(ii) If the diagonal of a square is 12 √2 cm then the perimeter of square is ……**

**(A) 24 cm **

**(B) 24 √2 cm **

**(C) 48 cm **

**(D) 48 √2 cm**

** **

**Solution:**

**In ∆ABC,**

**AC² = AB² + BC²**

**∴ (12 √2)² = AB² + AB² [□ABCD is a square]**

**∴ 2AB² = 12² × (√2)²**

**∴ 2AB² = 144 × 2**

**∴ 2AB² = 288**

**∴ AB² = \( \frac{288}{2} \) **

**∴ AB² = 144**

**∴ √(AB²) = √144 [Taking square root on both sides]**

**∴ AB = 12**

** **

**∴ Perimeter of □ABCD = 4 x 12**

**∴ Perimeter of □ABCD = 48 cm**

** **

**OPTION (C) – **48 cm

** **

**(iii) If opposite angles of a rhombus are (2x)⁰ and (3x – 40)⁰ then value of x is …**

**(A) 100 ⁰ **

**(B) 80 ⁰ **

**(C) 160 ⁰ **

**(D) 40 ⁰**

** **

**Solution:**

**2x = 3x – 40 [Opposite angles of a rhombus are congruent]**

**∴ 2x – 3x = –40**

**∴ -x = –40**

**∴ x = 40°**

**OPTION (D) **– 40°

**2. Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.**

**Given: **

**Let □ABCD be the rectangle.**

**∴ AB = 7 cm**

**and BC = 24 cm**

** **

**To find: **

**Length of its diagonal**

** **

**Solution:**

**In ∆ABC, **

**∠B = 90° [Angle of a rectangle]**

**∴ ∆ABC is a right angled triangle **

**∴ AC² = AB² + BC² [By Pythagoras theorem]**

**∴ AC² = 7² + 24²**

**∴ AC² = 49 + 576**

**∴ AC² = 625**

**∴ √(AC²) = √625 [Taking square root on both sides]**

**∴ AC = 25 cm**

** **

**Ans:** The length of the diagonal of the rectangle is 25 cm.

**3. If the diagonal of a square is 13 cm then find its side.**

**Given: **

**□PQRS and □MNRL are rectangles**

**M is the midpoint of side PR**

** **

**To prove:**

**SL = LR****LN = (\( \frac{1}{2} \)) SQ**

** **

**Proof:**

**□PQRS and □MNRL are rectangles [Given]**

**∴ ∠S ≅ ∠L ≅ ∠Q ≅ ∠N = 90° [Angles of rectangle]**

**∴ seg ML || seg PS …(i) [Corresponding angles test]**

**∴ seg MN || seg PQ …(ii) [Corresponding angles test]**

** **

**In ∆PRS,**

**Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]**

**∴ Point L is the midpoint of seg SR ……(ii) [Converse of midpoint theorem]**

**∴ SL = LR**

** **

**In ∆PRQ,**

**Point M is the midpoint of PR and seg MN || seg PQ [Given] [From (ii)]**

**∴ Point N is the midpoint of seg RQ ……(iii) [Converse of midpoint theorem]**

**∴ RN = NQ**

** **

**In ∆RSQ,**

**Points L and N are the midpoints of seg SR and seg QR respectively [From (ii) and (iii)]**

**∴ LN = \( \frac{1}{2} \) SQ [Midpoint theorem]**

** **

**Hence proved. **

**4. Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of each side.**

**Given: **

**Let □STUV be the parallelogram**

**Ratio of two adjacent sides of a parallelogram is 3 : 4**

**Perimeter of □STUV = 112 cm**

** **

**To find: **

**Length of each side**

** **

**Solution:**

**Let the common multiple be x.**

**∴ ST = 3x cm and TU ≅ 4x cm**

**∴ ST ≅ UV = 3x cm and TU = SV = 4x cm …..(i) [Opposite sides of a parallelogram]**

** **

**Perimeter of □STUV = 112 [Given]**

**∴ ST + TU + UV + SV = 112**

**∴ 3x + 4x + 3x + 4x = 112 [From (i)]**

**∴ 14x = 112**

**∴ x = \( \frac{112}{14} \) **

**∴ x = 8**

** **

**The measure of each sides are, **

**∴ ST ≅ UV = 3x = 3 x 8 = 24 cm [From (i)]**

**∴ TU ≅ SV = 4x = 4 x 8 = 32 cm [From (i)]**

** **

**Ans:** The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm

**5. Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.**

**Given: **

**□PQRS is a rhombus**

**PR = 20 cm**

**QS = 48 cm**

** **

**To find: **

**Length of side PQ**

** **

**Solution:**

**In □PQRS,**

**∴ PT = \( \frac{1}{2} \) PR [Diagonals of a rhombus bisect each other]**

**∴ PT = \( \frac{1}{2} \) x 20 **

**∴ PT = \( \frac{20}{2} \) **

**∴ PT = 10 cm**

** **

**Also, **

**QT = \( \frac{1}{2} \) QS [Diagonals of a rhombus bisect each other]**

**∴ QT = \( \frac{1}{2} \) x 48 **

**∴ QT = \( \frac{48}{2} \) **

**∴ QT = 24 cm**

** **

**In ∆PTQ, **

**∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]**

**∴ ∆PQT is a right angled triangle **

**∴ PQ² = PT² + QT² [By Pythagoras theorem]**

**∴ PQ² = 10² + 24²**

**∴ PQ² = 100 + 576**

**∴ PQ² = 676**

**∴ √(PQ²) = √676 [Taking square root of both sides]**

**∴ PQ = 26 cm**

** **

**Ans:** The length of side PQ is 26 cm.

**6. Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50⁰ then find the measure of ∠MPS.**

**Given: **

**□PQRS is a rectangle**

**∠QMR = 50⁰**

** **

**To find: **

**∠MPS**

** **

**Solution:**

**□PQRS is a rectangle [Given]**

**∴ PM = \( \frac{1}{2} \) PR …(i) [Diagonals of a rectangle bisect each other]**

**and MS = \( \frac{1}{2} \) QS …(ii) [Diagonals of a rectangle bisect each other]**

** **

**Also, **

**PR ≅ QS …..(iii) [Diagonals of a rectangle are congruent]**

** **

**∴ PM ≅ MS ….(iv) [From (i), (ii) and (iii)]**

** **

**In ∆PMS,**

**PM ≅ MS [From (iv)]**

**∴ ∠MSP = ∠MPS = x° …..(v) [Isosceles triangle theorem]**

** **

**∠QMR ≅ ∠PMS = 50° [Vertically opposite angles]**

**∵ ∠QMR = 50⁰**

**∴ ∠PMS = 50° ……(vi)**

** **

**In ∆MPS,**

**∠PMS + ∠MPS + ∠MSP = 180° [Sum of the measures of the angles of a triangle is 180°]**

**∴ 50° + x⁰ + x⁰ = 180° [From (v) and (vi)]**

**∴ 50° + 2x⁰ = 180**

**∴ 2x = 180 – 50**

**∴ 2x = 130**

**∴ x = \( \frac{130}{2} \) **

**∴ x = 65°**

** **

**∴ ∠MPS = x = 65° [From (v)]**

**Ans:** ∠MPS = 65°

**7. In the adjacent Figure 5.42, if ****seg AB || seg PQ , seg AB ≅ seg PQ, ****seg AC || seg PR, seg AC ≅ seg PR ****then prove that, seg BC || seg QR and seg BC ≅ seg QR.**

**Given: **

**seg AB || seg PQ **

**seg AB ≅ seg PQ**

**seg AC || seg PR**

**seg AC ≅ seg PR **

** **

**To prove: **

**seg BC || seg QR and seg BC ≅ seg QR**

** **

**Proof:**

**In □ABQP,**

**seg AB || seg PQ [Given]**

**seg AB ≅ seg PQ [Given]**

**∴ □ABQP is a parallelogram [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]**

**∴ segAP || segBQ …..(i)**

**∴ seg AP ≅ seg BQ …..(ii) [Opposite sides of a parallelogram are congruent]**

** **

**In □ACRP,**

**seg AC || seg PR [Given]**

**seg AC ≅ seg PR [Given]**

**∴ □ACRP is a parallelogram [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]**

**∴ seg AP || seg CR …(iii)**

**∴ seg AP ≅ seg CR …(iv) [Opposite sides of a parallelogram are congruent]**

** **

**In □BCRQ,**

**seg BQ || seg CR**

**seg BQ ≅ seg CR**

**∴ □BCRQ is a parallelogram [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]**

**∴ seg BC || seg QR**

**∴ seg BC ≅ seg QR [Opposite sides of a parallelogram are congruent]**

** **

**Hence proved. **

**8*. In Figure 5.43, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and ****PQ = \( \frac{1}{2} \) (AB + DC)**

**Given: **

**□ABCD is a trapezium**

**AB || DC**

**Point P is the midpoint of seg AD **

**Point Q is the midpoint of seg BC **

** **

**To prove: **

**PQ || AB and PQ = \( \frac{1}{2} \) (AB + DC)**

** **

**Construction:**

**Join points A and Q. Extend seg AQ and let it meet produced DC at R.**

** **

**Proof:**

**seg AB || seg DC and seg BC is their transversal [Given]**

**∴ ∠ABC ≅ ∠RCB [Alternate angles]**

**∴ ∠ABQ ≅ ∠RCQ ….(i) [B – Q – C]**

** **

**In ∆ABQ and ∆RCQ,**

**∠ABQ ≅∠RCQ [From (i)]**

**seg BQ ≅ seg CQ [Q is the midpoint of seg BC]**

**∠BQA ≅ ∠CQR [Vertically opposite angles]**

**∴ ∆ABQ ≅ ∆RCQ [ASA test]**

**seg AB ≅ seg CR …(ii) [c. s. c. t.]**

**seg AQ ≅ seg RQ [c. s. c. t.]**

**∴ Q is the midpoint of seg AR ….(iii)**

** **

**In ∆ADR,**

**Point P is the midpoint of seg AD and Point Q is the midpoint of seg AR [Given and from (iii)**

** **

**∴ seg PQ || seg DR [Midpoint theorem]**

**i.e. seg PQ || seg DC ……..(iv) [D-C-R]**

**But, seg AB || seg DC …….(v) [Given]**

**∴ seg PQ || seg AB [From (iv) and (v)]**

** **

**In ∆ADR,**

**PQ = \( \frac{1}{2} \) DR [Midpoint theorem]**

**PQ = \( \frac{1}{2} \) (DC + CR) [D – C – R]**

**PQ = \( \frac{1}{2} \) (DC + AB)**

**i.e. PQ = \( \frac{1}{2} \) (AB + DC)**

**Hence Proved.**

**9. In the adjacent figure 5.44, □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.**

**Given: **

**□ABCD is a trapezium **

**AB || DC**

**Point M is the midpoint of diagonal AC Point N is the midpoint of diagonal DB**

** **

**To prove: **

**MN || AB**

** **

**Construction: **

**Join D and M. Extend seg DM to meet seg AB at point E such that A – E – B**

** **

**Proof:**

**seg AB || seg DC and seg AC is their transversal [Given]**

**∴ ∠CAB ≅ ∠ACD [Alternate angles]**

**∴ ∠MAE ≅ ∠MCD ….(i) [C – M – A, A – E – B]**

** **

**In ∆AME and ∆CMD, **

**∠AME ≅ ∠CMD [Vertically opposite angles]**

**seg AM ≅ seg CM [M is the midpoint of seg AC]**

**∠MAE ≅∠MCD [From (i)]**

**∴ ∆AME ≅ ∆CMD [ASA test]**

**∴ seg ME ≅ seg MD [c.s.c.t]**

**∴ Point M is the midpoint of seg DE. …(ii)**

** **

**In ∆DEB,**

**Point M is the midpoint of diagonal DE and Point N is the midpoint of diagonal DB [Given and from (ii)]**

**∴ seg MN || seg EB [Midpoint theorem]**

**∴ seg MN || seg AB [A – E – B]**

** **

**Hence proved. **