Question

In the adjacent figure $ABCD$ is a square and $\Delta APB$ is an equilateral triangle. Prove that $\Delta APD\cong \Delta BPC$ .
(Hint: In $\Delta APD$ and $\Delta BPC\overline{AD}=\overline{BC},\overline{AP}=\overline{BP}$ and
$\angle PAD=\angle PBC={90}^{\circ }-{60}^{\circ }={30}^{\circ })$

Answer 1

Given $ABCD$ is square

$\therefore AB=BC=CD=AD$

$\angle A=\angle B=\angle C=\angle D={90}^0$

Given $PDB$ is equilateral

$\therefore AB=PD=PB$

$\angle \left(PAB\right)=\angle \left(APB\right)=\angle \left(PBA\right)={60}^0$

$\therefore \angle \left(DAP\right)=\angle \left(DAB\right)-\angle \left(PAB\right)$

$=90-60={30}^0$

$\therefore \angle \left(CBP\right)=\angle \left(CBA\right)-\angle \left(PBD\right)$

$=90-60={30}^0$

In $\Delta le$ $DPA$ and $CPB$

$AD=BC$ [sides of the square]

$PA=PB$ [sides of the equilateral]

$\angle \left(DAP\right)=\angle \left(CBP\right)={60}^0$

$\therefore \Delta DPA\equiv \Delta CPB$ [SAS axion]

$\Rightarrow \Delta DPA\cong \Delta BPC$

Hence proved.