Question

In the adjoining figure, $△ABC$ is right angled at $B$ and $P$ is the mid-point of $AC$. Show that, $PA=PB=PC$.

Answer 1

In triangle $\Delta PBQ$ and $\Delta PAQ$

$\angle PBA=\angle PBC$

$\angle PQB=\angle PBC={90}^{\circ }$ (Each angles is a right angle.)

$PQ=PQ$ (Taking Common in both triangle)

Using a Right Hand Congruence rule:

$\Delta PBQ\cong \Delta PAQ$

$⟹PA=PB\dots \left(1\right)$

Similarly, if $R$ is the mid-point of $BC$ and $RC$ is formed, then in triangle $\Delta PBR$ and $\Delta PCR$:

$\angle PRB=\angle PRC$ (Each angles is a right angle.)

$BR=CR$ ($R$ is the mid-point)

$PR=PR$

Using a Right Hand congruence rule:

$\Delta PBR\cong \Delta PCR$

$PB=PC\dots \left(2\right)$

Hence, equating eq $\left(1\right)$ and $\left(2\right)$ we get:

$PA=PB=PC$

Hence, this is the required answer.