Question

In the adjoining figure, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet DC at P, prove that (i) ∠APB = 90°, (ii) AD = DP and PB = PC = BC, (iii) DC = 2AD.

Answer 1

Given ABCD is a parallelogram.
So, opposite sides are equal and sum of adjacent angles equals to ${180}^{\circ }$
Thus, $\angle A=\angle C={60}^{\circ }$ and $\angle B=\angle D$ and also
$\Rightarrow \angle A+\angle B=180$
$\Rightarrow \angle B=180-60$
$\therefore \angle B={120}^{\circ }$
Thus, $\angle B=\angle D={120}^{\circ }$
(i) To prove $\angle APB={90}^{\circ }$:
In a triangle $APB$, $\angle APB+\angle PBA+\angle BAP={180}^{\circ }$
$\angle APB+60+30=180$ as AP and BP are bisectors of $\angle A$ and $\angle B$ respectively.
$\angle APB=180-90$
$\therefore \angle APB={90}^{\circ }$
(ii) To prove AD=DP and PB=PC=BC:
Consider triangle $\Delta ADP$,
So, $\angle ADP+\angle DPA+\angle PAD={180}^{\circ }$
$\Rightarrow 120+\angle DPA+30=180$
$\Rightarrow \angle DPA=180-150$
$\therefore \angle DPA={30}^{\circ }$ and $\angle PAD=30$
So, triangle $\Delta ADP$ forms an isosceles triangle with equal sides AD=DP.
Similar way consider triangle $\Delta PBC$,
We have $\angle PBC={60}^{\circ },\angle CPB={180}^{\circ }-{(90+30)}^{\circ }=(180-120{)}^{\circ }={60}^{\circ }$
Thus, $\angle PCB={60}^{\circ }$
$\angle PBC=\angle CPB=\angle PCB$
Therefore, $\Delta PBC$ is an equilateral triangle.
So, PB=PC=BC.
(iii) To prove DC=2AD
Since DC=DP+PC
$DC=AD+BC$ [Since, AD=DP and PC=BC]
As opposite sides AD and BC are equal in parallelogram.
So, $DC=AD+AD$
$\therefore DC=2AD$