Question

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

In each case,$\Delta APD\cong \Delta BPC$.

State whether the above statement is true or false.

• A. True
• B. False

Answer 1

The correct option is A True

$\angle A=\angle B={90}^{\circ }$ (ABCD is a square)

$\angle PAB=\angle PBC={60}^{\circ }$ (APB is an equilateral triangle)

$\angle A-\angle PAB=\angle B-\angle PBC$

$\angle PAD=\angle PBC$ (I)

Now, in $△PAD$ and $△PBC$,
$AD=BC$ (Opposite sides of the square ABCD)

$\angle PAD=\angle PBC$ (From I)

$PA=PB$ ($△PAB$ is an equilateral triangle)

Hence, $△PAD\cong △PBC$ (SAS Postulate)

$⟹\Delta APD\cong \Delta BPC$.