Question

In any triangle $ABC$, if the angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect, prove that they intersect on the circumcircle of the $△ABC$.

Answer 1

Step $1$: Find the relation between $AP$ and $PE$.

Let Angle bisector of $\angle A$ and Perpendicular bisector of $BC$ intersect at $E$.

Since, the perpendicular bisector of a chord always passes through the center. Thus, the Perpendicular bisector of $BC$ will pass through the center of the circle $O$.

$\Rightarrow OE$ is a straight line.

Let's do the Construction:

1. Join $OA$. $O$ is the center of the circle.
2. Draw a perpendicular $OP\perp AE$.

Since, Perpendicular from the center bisects the chord. Thus $AP=PE$……..$\left(1\right)$

Step $2$: Prove the statement.

In $△AOP$ and $△\mathrm{POE}$

$\angle OPA=\angle OPE$ $[\because$Construction$]$

$OP=OP$ $[\because$Common side$]$

$AP=PE$ $[\because$Equation $\left(1\right)$$]$

By $\mathrm{RHS}$ rule, $△AOP\cong △POE$

$\Rightarrow OA=OE$ $\left[\because △AOP\cong △POE\right]$

Since $\mathrm{OA}=\mathrm{OE}=$Radius of circle.

Therefore, $E$ lies on the circumcircle of $△ABC$.

Hence proved that the angle bisector of $\angle A$ and perpendicular bisector of $BC$ intersect on the circumcircle of the $△ABC$.