Question

Triangle ABC is an equilateral triangle . the bisector of angle B intersects circumcircle of triangle ABC at P prove that CQ=CA

Answer 1

Answer:

Angles in a semicircle is a right angle.

Answer 2

Use Converse of Isosceles triangle Theorem

Step-by-step explanation:

Given,

ΔABC is an equilateral triangle.

BP is the bisector of $\angle B$

• To Prove: CQ = CA  

• Proof: ΔABC is an equilateral triangle. (Given)

$\angle ABC + \angle CAB + \angle ACB = 60$° (Angles of Equilateral Triangle are all equal, 60° each)

$\angle ABP = \angle CBP$ (BP is Bisector)

$\angle CBP = \frac {1}{2} \times \angle ABC$

$\angle CBP = \frac {1}{2} \times$ 60° = 30°

$\angle CBP = \angle CAP$ = 30° """(1) (Angles incircled in the same arc)

$\angle ACB$ = 60° (Given)

So, $\angle ACQ$ = 180° - 60° = 120° (Linear pair)

Therefore, $\angle ACQ$ = 120° """(2)

In ΔACQ,

Therefore, $\angle AQC$ = 180° - (30° + 120°) From (1) & (2)

⇒ $\angle AQC$ = 180° - 150°

⇒ $\angle AQC$ = 30° ""(3)

In ΔACQ, $\angle CAQ = \angle AQC$ = 30 °

From (1) and (3)  

Therefore, CQ = CA (Converse of Isosceles triangle Theorem)