Question

In the given fig. ABCD is a square and $△PAB$ is an equilateral triangle.
(i) Prove that $△APD\cong △BPC$
(ii) Show that $\angle DPC=30$

Answer 1

In $△ABP$,

$AP=BP$ [Sides of equilateral $△$]

& $AD=BC$ [Sides of equilateral $△$]

and $\angle DAP=\angle DAB-\angle PAB$$={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

Now in $△APD$ and $△BPC$

$AP=BP$ [Sides of equilateral $△$]

$AD=BC$ [Sides of equilateral $△$]

$\angle DAP=\angle CBP$ [Both ${30}^{\circ }$]

$\therefore △APD\cong △BPC$ [By SAS]

In $△APD$

$AP=AD$ [$\because$ AP=AB ]

$\angle DAP={30}^{\circ }$

$\angle APD=\frac{{180}^{\circ }-{30}^{\circ }}{2}={75}^{\circ }$

Similarly, $\angle BPC={75}^{\circ }$

Therefore, $\angle DPC={360}^{\circ }-(75+75+60)={150}^{\circ }$