Question

In the following diagrams, ABCD is a square and APB is an equilateral triangle.

In each case,

(i) Prove that : $\Delta$ APD $\cong \Delta$ BPC

(ii) Find the angles of $\Delta$ DPC.

Answer 1

Solution :

Part 1
Given $△ABP$ is an equilateral triangle.
Then, AP = BP (Same side of the triangle)
and AD = BC (Same side of square)
and, $\angle DAP=\angle DAB-\angle PAB={90}^0-{60}^0={30}^0$
Similarly,
$\angle BPC=\angle ABC-\angle ABP={90}^0-{60}^0={30}^0$
$\therefore \angle DAP=\angle BPC$

$\therefore △APD\cong to△BPC$ ( SAS proved)
2nd part
In $△APD$
AP = AD II as AP = AB (equilateral triangle)
We know that $\angle DAP={30}^0$
$\therefore \angle APD=\frac{(180°-30°)}{2}$

$\because △APD$ is an isosceles
triangle and $\angle APD$ is one of the base angles.
$=\frac{{150}^0}{2}={75}^0$
$=\angle APD={75}^0$
Similarly, $\angle BPC={75}^0$
$\therefore \angle DPC={360}^0-({75}^0+{75}^0+{60}^0)$
= $\angle DPC={150}^0$
Now in $△PDC$
PD = PC as $△APDis\cong △BPC$
$\therefore △PDC$is an isosceles triangle
And $△PDC=△PCD=({180}^0-\frac{{150}^0}{2})$or $\angle PDC=\angle PCD={15}^0$
$\angle DPC={150}^0$

$\angle PDC={15}^0$

and $\angle PCD={15}^0$