Q

# In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that: (iii) ∆ AQB ≅ ∆ CPD

Q : 9     In parallelogram ABCD, two points P and Q are taken on diagonal BD such that  $\small DP=BQ$ (see Fig. $\small 8.20$). Show that:

(iii) $\small \Delta AQB\cong \Delta CPD$

Views

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$.

To prove :$\small \Delta AQB\cong \Delta CPD$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ       (Given )

$\angle$ABQ=$\angle$CDP     (alternate angles)

AB=CD       (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$                (By SAS)

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