Question

In the adjacent figure, $AB\parallel QP\parallel CD$, Q is the midpoint of AC. If AB=4cm and CD=6cm then find PQ.

Answer 1

$Q$ is midpoint of $AC$

Since, $PQ\parallel CP$

$\therefore AP=PR$

Also $PQ=\frac{1}{2}CR---------\left(1\right)$

In $△ABP$ and $△RPD$

$\angle BAP=\angle DRP$(alternate angle )

$AP=PR$

$\angle APB=\angle RPD$ (vertically angle )

$△ABP$ is congruent to $△RPD\left(ASA\right)$

So, $DR=AB=4cm$ (By $CPCT$)

We know that

$CR=CD+DR$

$CR=CD+AB----------\left(2\right)$

from eq $\left(1\right)$ and $\left(2\right)$

$PQ=\frac{1}{2}(CD+AB)$

$PQ=\frac{1}{2}(6+4)$

$PQ=5cm$