Question

In the below-shown figure if AP=BP, AQ=BQ, then can we say that AM=BM?

• A. True
• B. False

Answer 1

The correct option is A

True

In the given figure,
PA=PB (Given)
AQ=BQ (Given)
PQ is common.
So, $\Delta$ PAQ $\cong$ $\Delta$ PBQ (SSS rule).
$\Rightarrow \angle$ APQ=$\angle$ BPQ (CPCT).
Also, PA=PB (Given) and PM is common
Hence, $△$ PMA $\cong$ $△$PMB (SAS Rule)

So, AM=BM and $\angle$PMA = $\angle$PMB (CPCT)

$\angle$PMA + $\angle$PMB = ${180}^{\circ }$ (Linear pair axiom),

$\angle$ PMA = $\angle$ PMB = ${90}^{\circ }$

Therefore, PM, that is, PMQ is the perpendicular bisector of AB.