maths

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

In each case,$ΔAPD≅ΔBPC$.

**State whether the above statement is true or false.**

$∠PAB=∠PBC=60_{∘}$ (APB is an equilateral triangle)

$∠A−∠PAB=∠B−∠PBC$

$∠PAD=∠PBC$ (I)

Now, in $△PAD$ and $△PBC$,

$AD=BC$ (Opposite sides of the square ABCD)

$AD=BC$ (Opposite sides of the square ABCD)

$∠PAD=∠PBC$ (From I)

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

Hence, $△PAD≅△PBC$ (SAS Postulate)

$⟹ΔAPD≅ΔBPC$.

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