In the given figure, $A P = D P$ , $∠ A = ∠ D$ and $∠ B = ∠ C = 90^{o}$
which of the following is correct

P

is the mid-point of

B C

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△ A B P ≅ △ D C P

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A B = D C

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A B = P B

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Solution

The correct option is C $A B = D C$
In triangles $△ A B P$ and $△ D C P$
$∠ B = ∠ C$ (given)
$∠ A = ∠ D$ (given)
$A P = D P$ (given)
Hence, $△ A B P ≅ △ D C P$ (AAS)
So,
$B P = C P$ (C.P.C.T)
so, $P$ is the mid-point of $B C$
Also, $A B = D C$ (C.P.C.T)
Hence, option $(a)$ , $(b)$ and $(c)$ are correct.

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