Question

In the figure , ABCD is a parallelogram in which $\angle A={60}^{\circ }$. If the bisectors of $\angle Aand\angle B$ meet at P, prove that AD =DP, PC =BC and DC = 2AD.

Answer 1

In || gm ABCD,

$\angle A={60}^{\circ }$

Bisector of $\angle Aand\angle B$ meet at P.

To prove :

(i) AD = DP

(ii) PC = BC

(iii) DC = 2AD

Construction : Join PD and PC

Proof : In ||gm ABCD,

$\angle A={60}^{\circ }$

But $\angle A+\angle B={180}^{\circ }$ (Sum of excutive angles)

$\Rightarrow {60}^{\circ }+\angle B={180}^{\circ }$

$\angle B={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

$\because DC||AB$

$\therefore \angle PAB=\angle DPA$ (alternate angles)

$\Rightarrow \angle PAD=\angle DPA$ ($\because \angle PAB=\angle PAD)$

$\therefore$ AD = DP (PA is its angle bisector , sides opposite to equal angles)

(ii) Similarly, we can prove that

$\angle PBC=\angle CPB$ ($\because \angle PAB=\angle BCAalternateangles)$

$\therefore$ PC =BC

(iii) DC = DP + PC

=AD+BC [From (i) and (ii)]

= 2AD

$(\because$ BC =AD opposite sides of the ||gm)

Hence DC = 2AD