Question

how to prove it?
plz give answer​

Answer 1

ΔABC is an equilateral triangle.

∴ ∠ABC = ∠BCA = ∠BAC = 60°

BP is the angle bisector of ∠ABC.

∴ ∠ABP = ∠CBP = 60° / 2 = 30°

Here,

∠CBP = ∠CAP = 30° (Both are angles on the same arc of the circle)

∠BCA and ∠QCA are linear pairs.

∠QCA = 180° - 60° = 120°

In ΔQCA,

∠C = 120°

∠A = 30°

∴ ∠Q = 180° - (120° + 30°) = 180° - 150° = 30°

∴ ΔQCA is an isosceles triangle.

∠A = ∠Q

∴ CQ = CA

Hence, proved!!!

Thank you. Have a nice day. :-)

#adithyasajeevan

Answer 2

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HERE IS YOUR ANSWER. :-




$\huge{ \green{ \:answer}} >$Since AB = AC, hence angle opposite to equal sides are also equal.



< ACB = < ABC.



Since, ext < ACB = < QAC + < AQC



Then, < ABC = < QAC + < AQC



Now, because < ABP = < PBC.




Therefore,
2 < PBC = < ABC





Hence, 2 < PBC = < QAC + < AQC






Now,
< QAC = < PAC and because ∆PAC and ∆PBC lie on same line segment PC.






So,
< PAC = < PBC

Hence < PBC = < QAC



2 < PBC = < PBC + AQC



< PBC = < AQC


< PAC = < AQC


< QAC = < AQC

Side opposite to equal sides are equal.

QC = AC

Hence prove.








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