**Telanagana SCERT Solution Class IX (9) Math Chapter 8 Quadrilaterals Exercise 8.2**

**(1) In the adjacent figure ABCD is a parallelogram and ABEF is a rectangle show that ΔAFD ****≅**** ΔBEC**

**Solution:** Given,

ABCD is a parallelogram

ABEF is a rectangle

∴ In △AFD & △BEC

(i) ∠AFD = 90^{o} [angle of rectangle ABEF]

∠BEF = 90^{o} [angle of a rectangle ABEF]

∴ ∠BEF + ∠BEC = 180^{o}

= 90^{o}

∴ ∠AFD = ∠BEC = 90^{o}

(ii) AF = BE [opposite side of rectangle ABEF]

(iii) AD = BC [opposite side of parallelogram ABCD]

Now,

∴ △AFD is a right angled triangle at ∠F = 90^{o}

△BEC is aright angled triangle at ∠E = 90^{o} with BC as hypotenuse

∴ △AFD ≅ △BEC by RHS

**(2) Show that the diagonals of a rhombus divide it into four congruent triangles.**

**Solution:** ABCD is a rhombus

AC and BD are its diagonals

In △AOD & △BOC

(i) ∠AOD = ∠BOC [vertically opposite angles]

(ii) AO = OC

(iii) DO = DB [∵ diagonals of a rhombus bisects each other]

∴ △AOD ≅ △BOC by SAS

Similarly,

△AOB ≅ △DOC by SAS

Now,

In △AOD and △DOC

(i) AD = DC [∵ all sides of a rhombus are equal]

(ii) ∠AOD = ∠DOC = 90^{o} [∵ diagonals of a rhombus intersects each other at right angles]

(iii) DO is the common side

Hence,

△AOD & △DOC are right angled triangle with AD and DC being their hypotenuses respectively.

∴ △AOD ≅ △DOC by RHS

Similarly,

△BOC ≅ △AOB by RHS

∴ △AOD ≅ △AOB ≅ △BOC ≅ △DOC

∴ We can conclude that diagonals of a rhombus divided it into four congruent triangles

**(3) In a quadrilateral ABCD, the bisector of ****∠****C and ****∠****D intersect at O**

**Solution:** Given,

ABCD is a quadrilateral

DO and CO are bisects of ∠D and ∠C respectively

Now,

In △DOC

△COD = 180^{o} – (∠DOC + ∠OCD)

= 180^{o} – ( +(∠C)/2) [DO and CO are bisects of D and C]

= 180^{o} – ½(∠C+ ∠D)

Or, COD = ½ 360^{o} – (∠C+ ∠D)

Or, 2COD = 360^{o} – (∠C+ ∠D)

Or, ∠C+ ∠D = 360^{o} – 2∠COD – (i)

Now, We know

Sum of angles of a quadrilateral is 360^{o}

∠A + ∠B + ∠C + ∠D = 360^{o}

Putting the value of ∠C + ∠D from equation (i) we get

∠A + ∠B + 360^{o} – 2∠COD = 360^{o}

Or, ∠A + ∠B – 2∠COD = 0

Or, 2∠COD = ∠A + ∠B

Or, ∠COD = ½(∠A + ∠B) Hence proved