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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.
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Solution
∆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]
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