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]