Question

In the following figure $\Delta ABC$ is an equilateral triangle. Bisector of $\angle B$ intersects circumcircle of $\Delta ABC$ in point $P$. Prove that: $CQ=CA$

Answer 1

Given: $\Delta ABC$ is an equilateral triangle.
BP is the bisector of $\angle B$.
To Prove : CQ = CA
Proof : $\Delta ABC$ is an equilateral $\Delta$ (Given)
$\angle ABC=\angle CAB=\angle ACB={60}^o$ (Given)
$\angle ABP=\angle CBP$ (BP is the bisector)
$\angle CBP=\frac{1}{2}\times \angle ABC$
$\angle CBP=\frac{1}{2}\times {60}^o={30}^o$
$\angle CBP=\angle CAP={30}^o$ ..... (1) ($\angle$s incircled in the same arc)
$\angle ACB={60}^o$ (Given)
$\therefore \angle ACQ={180}^o-{60}^o={120}^o$ (Linear pair)
$\therefore \angle ACQ={120}^o$ ..... (2)
In $\Delta ACQ$,
$\therefore \angle AQC={180}^o-({30}^o+{120}^o)$ From (1) and (2)
$\angle AQC={180}^o-{150}^o$
$\therefore \angle AQC={30}^o$ ...... (3)
In $\Delta ACQ$, $\angle CAQ=\angle AQC={30}^o$
From (1) and (3)
$\therefore CQ=CA$ (Converse of isosceles $\Delta$ theorem)