**Chapter 5 Quadrilaterals Practice Set 5.2**

Chapter 5 Quadrilaterals Practice Set 5.2

**Question 1.In the adjoining figure, â–¡ABCD is a parallelogram, P and Q are midpoints of sides AB and DC respectively, then prove â–¡APCQ is a parallelogram.**

Given: â–¡ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively.

To prove: â–¡APCQ is a parallelogram.

Solution:

Given: â–¡ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively.

To prove: â–¡APCQ is a parallelogram.

Solution:

âˆ´ AP = QC â€¦.(iii) [From (i) and (ii)]

Also, AB || DC [Opposite angles of a parallelogram]

i.e. AP || QC â€¦.(iv) [A â€“ P â€“ B, D â€“ Q â€“ C]

From (iii) and (iv),

â–¡APCQ is a parallelogram. [A quadrilateral is a parallelogram if its opposite sides is parallel and congruent]

**Question 2.Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.Given:â–¡ABCD is a rectangle.To prove: Rectangle ABCD is a parallelogram.Solution:**

Proof:

â–¡ABCD is a rectangle.

âˆ´ âˆ A â‰… âˆ C = 90Â° [Given]

âˆ B â‰… âˆ D = 90Â° [Angles of a rectangle]

âˆ´ Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

**Question 3.In the adjoining figure, G is the point of concurrence of medians of ADEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that â–¡GEHF is a parallelogram.Given: Point G (centroid) is the point of concurrence of the medians of ADEF.DG = GHTo prove: â–¡GEHF is a parallelogram.Solution:**

Proof:

Let ray DH intersect seg EF at point I such that E-I-F.

âˆ´ seg DI is the median of âˆ†DEF.

âˆ´ El = FI â€¦â€¦(i)

Point G is the centroid of âˆ†DEF.

âˆ´ DG = 2(GI)

âˆ´ GH = 2(GI) [DG = GH]

âˆ´ GI + HI = 2(GI) [G-I-H]

âˆ´ HI = 2(GI) â€“ GI

âˆ´ HI = GI â€¦.(ii)

From (i) and (ii),

â–¡GEHF is a parallelogram [A quadrilateral is a parallelogram, if its diagonals bisect each other]

**Question 4.Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.Given: â–¡ABCD is a parallelogram.Rays AS, BQ, CQ and DS bisect âˆ A, âˆ B, âˆ C and âˆ D respectively.To prove: â–¡PQRS is a rectangle.Solution:**

Proof:

âˆ BAS = âˆ DAS = xÂ° â€¦(i) [ray AS bisects âˆ A]

âˆ ABQ = âˆ CBQ =yÂ° â€¦.(ii) [ray BQ bisects âˆ B]

âˆ BCQ = âˆ DCQ = uÂ° â€¦..(iii) [ray CQ bisects âˆ C]

âˆ ADS = âˆ CDS = vÂ° â€¦.(iv) [ray DS bisects âˆ D]

â–¡ABCD is a parallelogram. [Given]

âˆ´ âˆ A + âˆ B = 180Â° [Adjacent angles of a parallelogram are supplementary]

âˆ´ âˆ BAS + âˆ DAS + âˆ ABQ + âˆ CBQ = 180Â° [Angle addition property]

âˆ´ xÂ°+xÂ°+ vÂ° + vÂ° = 180 [From (i) and (ii)]

âˆ´ 2xÂ° + 2vÂ° =180

âˆ´ x + y = 90Â° â€¦â€¦(v) [Dividing both sides by 2]

Also, âˆ A + âˆ D= 180Â° [Adjacent angles of a parallelogram are supplementary]

âˆ´ âˆ BAS + âˆ DAS + ADS + âˆ CDS = 180Â° [Angle addition property]

âˆ´ xÂ° + xÂ° + vÂ° + vÂ° = 180Â°

âˆ´ 2xÂ° + 2vÂ° = 180Â°

âˆ´ xÂ° + vÂ° = 90Â° â€¦..(vi) [Dividing both sides by 2]

In âˆ†ARB,

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

âˆ´ xÂ° + yÂ° + âˆ SRQ = 180Â° [A â€“ S â€“ R, B â€“ Q â€“ R]

âˆ´ 90Â° + âˆ SRQ = 180Â° [From (v)]

âˆ´ âˆ SRQ = 180Â°- 90Â° = 90Â° â€¦..(vi)

Similarly, we can prove

âˆ SPQ = 90Â° â€¦(viii)

In âˆ†ASD,

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

âˆ´ âˆ ASD + xÂ° + vÂ° = 180Â° [From (vi)]

âˆ´ âˆ ASD + 90Â° = 180Â°

âˆ´âˆ ASD = 180Â°- 90Â° = 90Â°

âˆ´ âˆ PSR = âˆ ASD [Vertically opposite angles]

âˆ´ âˆ PSR = 90Â° â€¦..(ix)

Similarly we can prove

âˆ PQR = 90Â° ..(x)

âˆ´ In â–¡PQRS,

âˆ SRQ = âˆ SPQ = âˆ PSR = âˆ PQR = 90Â° [From (vii), (viii), (ix), (x)]

âˆ´ â–¡PQRS is a rectangle. [Each angle is of measure 90Â°]

**Question 5.In the adjoining figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS, then prove that â–¡PQRS is a parallelogram.Given: â–¡ABCD is a parallelogram.AP = BQ = CR = DSTo prove: â–¡PQRS is a parallelogram.Solution:**

Proof:

â–¡ABCD is a parallelogram. [Given]

âˆ´ âˆ B = âˆ D â€¦.(i) [Opposite angles of a parallelogram]

Also, AB = CD [Opposite sides of a parallelogram]

âˆ´ AP + BP = DR + CR [A-P-B, D-R-C]

âˆ´ AP + BP = DR + AP [AP = CR]

âˆ´ BP = DR â€¦.(ii)

In APBQ and ARDS,

seg BP â‰… seg DR [From (ii)]

âˆ PBQ â‰… âˆ RDS [From (i)]

seg BQ â‰… seg DS [Given]

âˆ´ âˆ†PBQ â‰… âˆ†RDS [SAS test]

âˆ´ seg PQ â‰… seg RS â€¦..(iii) [c.s.c.t]

Similarly, we can prove that

âˆ†PAS â‰… âˆ†RCQ

âˆ´ seg PS â‰… seg RQ â€¦.(iv) [c.s.c.t]

From (iii) and (iv),

â–¡PQRS is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

**Intext Questions and Activities**

**Question 1.Points D and E are the midpoints of side AB and side AC of âˆ†ABC respectively. Point F is on ray ED such that ED = DF. Prove that â–¡AFBE is a parallelogram. For this example write â€˜givenâ€™ and â€˜to proveâ€™ and complete the proof. (Text book pg. no. 66)Given: D and E are the midpoints of side AB and side AC respectively.ED = DFTo prove: â–¡AFBE is a parallelogram.Solution:**

Proof:

seg AB and seg EF are the diagonals of â–¡AFBE.

seg AD â‰… seg DB [Given]

seg DE â‰… seg DF [Given]

âˆ´ Diagonals of â–¡AFBE bisect each other.

âˆ´ â–¡AFBE is a parallelogram. [ By test of parallelogram]