In parallelogram $A B C D$ , two points $P$ and $Q$ are taken on diagonal $B D$ such that $D P = B Q$ . Show that:
$Δ A P D ≅ Δ C Q B$

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Solution

From the figure we get,
$\begin{matrix} A D ∥ B C a n d t r a n s v e r s a l B D ∠ A D P = ∠ C B Q (a l t e r n a t e a n g l e) - - - (1) I n Δ A P D a n d Δ C Q B A D = C B (O p p o s i t e s i d e o f p a r a l l e log r a m a r e e q u a l) ∠ A D P = ∠ C B Q [f r o m (1)] D P = B Q (g i v e n) ∴ Δ A P D ≅ Δ C Q B p r o v e d . \end{matrix}$

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Δ A P D ≅ Δ C Q B

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Δ

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In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure).

Show that:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

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