Question

In the figure, ∆ABC is an equilateral triangle .The angle bisector of ∠B will intersect the
circumcircle ∆ABC at point P.Then prove that CQ=CA.

Answer 1

Given:

• ∆ABC is an equilateral triangle.
• BP is the bisector of $\sf\angle$B

To Prove:

• CQ=CA

Proof:

∆ABC is an equilateral triangle. (Given)

$\sf\angle$ABC = $\sf\angle$CAB = $\sf\angle$ACB = 60°. (Given)

$\sf\angle$ABP = $\sf\angle$CBP (BP is the bisector)

$\sf\angle$CBP = 1/2 × $\sf\angle$ABC

$\sf\angle$CBP = 1/2 × 60° = 30°

$\sf\angle$CBP = $\sf\angle$CAP = 30°-----(i) ( $\sf\angle$s incircled in the same arc)

$\sf\angle$ACB = 60°. (Given)

$\therefore$ $\sf\angle$ACQ = 180°-60° = 120° (Linear Pair)

$\therefore$ $\sf\angle$ACQ = 120° -------(2)

In ∆ACQ,

$\therefore$ $\sf\angle$AQC = 180°-(30°+120°). from(1)and(2)

$\sf\angle$AQC = 180° - 150°

$\therefore$ $\sf\angle$AQC = 30°. -----(3)

In ∆ACQ,

$\sf\angle$CAQ= $\sf\angle$AQC = 30°. from (1) and (2)

$\therefore$CQ = CA. (converse of isosceles ∆theorem)