In the given figure, ABCD is a square, M is the midpoint of AB and PQ ⊥ CM meets AD at P and CB produced at Q. Prove that (i) PA = BQ, (ii) CP = AB + PA.

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Solution

Proof:
$In △ AMP and △ BMQ, we have : MA = MB (M is midpoint) ∠ AMP = ∠ BMQ (Vertically opposite angle s) ∠ PAM = ∠ QBM (90 ° each) ∴ △ AMP ≅ △ BMQ (ASA criterion)$
$\Rightarrow P A = B Q (CPCT) . . . (i)$
and $M P = M Q$ (CPCT)

$Now, in △ CMP and △ CMQ, we have : CM = CM (Common) ∠ CMP = ∠ CMQ (90 ° each) MP = MQ (Proved) ∴ △ CMP ≅ △ CMQ (SAS criterion)$

Thus, $CP = CQ = CB + BQ (Corresponding sides of congruent triangles)$
But $C B = A B (Sides of a square)$
$B Q = P A [From (i)]$
∴ $C P = A B + P A$
Hence, proved.

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