Question

In figure AN and CP are perpendiculars to the diagonal BD of a parallelogram ABCD. Prove that : $\left(i\right)\Delta ADN\cong \Delta CBP$ $\left(ii\right)AN=CP$

Answer 1

Since ABCD is a parallelogram.
$\therefore AD||BC$
Now, AD || BC and transversal BD intersects them at D and B.
$\therefore \angle 1=\angle 2$ [Alternate interior angles are equal]
Now, in $\Delta ADN$and $\Delta CBP$, we have
$\angle 1=\angle 2$
$\angle AND=\angle CPB$(right angle)
AD = BC [Opposite sides of a || gm are equal]
So, by AAS criterion of congruence
$\Delta ADN\cong \Delta CBP$
$AN=CP$
[Corresponding parts of congruent triangles are equal]
Hence proved.