Question

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

Answer 1

∆ABC is an equilateral triangle.

∴ ∠ABC = ∠ACB = ∠BAC = 60°   ...(i) [Angles of an equilateral triangle]

`∠CBP = 1/2 ∠ABC`   ...[Ray BP bisects ∠B]

∴ `∠CBP = 1/2 xx 60^circ`   ...[From (i)]

∴ ∠CBP = 30°

∠CBP = ∠CAP = 30°   ...[Angles inscribed in the same arc]

∴ ∠CAQ = 30°   ...(ii) [A–P–Q]

In ∆ABQ,

∠BAQ = ∠BAC + ∠CAQ   ...[Angle addition property]

∴ ∠BAQ = 60° + 30°   ...[From (i) and (ii)]

∴ ∠BAQ = 90°

Also, ∠ABQ = 60°   ...[From (i) and B–C–Q]

∴ ∠BQA = 30°   ...[Remaining angle of ∆ABQ]

∴ ∠CQA = 30°   ...(iii) [B–C–Q]

In ∆CQA,

∠CAQ = ∠CQA   ...[From (ii) and (iii)]

∴ CQ = CA   ...[Converse of isosceles triangle theorem]