Question

Given :- chord AB=chord BC=chord AC, angle BAP = angle CAP.

TO PROVE :- Seg CB =Seg CQ​

Answer 1

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Since it is given, chord AB = chord BC = chord AC

∴ ∆ ABC will be an equilateral triangle so ∠CAB = ∠ABC = ∠ACB = 60°

Also given, ∠BAP = ∠CAP = ½ * ∠CAB = 30°

∠ACB + ∠BCQ = 180° [Linear pair]

> 60° + ∠BCQ = 180°

> ∠BCQ = 180° - 60°

> ∠BCQ =120°

Therefore, we can say

∠AQB = ½ * [arcAB – arcCP] = ½ * [120° - 60°] = ½ * 60° = 30°

∠AQB + ∠CBQ + ∠BCQ = 180°

⇒ 30° + angle CBQ + 120° = 180°

⇒ ∠CBQ = 180° - (120° + 30°)

⇒ ∠CBQ = 30°

• Since in ∆CBQ, ∠CBQ = ∠AQB = 30°

∴ ∆CBQ is an isosceles triangle

⇒ seg CB = seg CQ

Answer 2

Step-by-step explanation:

• Given data

        AB =BC =CA

        So ΔABC is an equilateral triangle.

• We can also write

        ∠ACB = 60°    (All angle of equilateral triangle is 60°)

        ∠BAC = 60°

• ∠BAP =∠PAC = 30°     (Given ∠BAP =∠PAC)

        Or

        ∠BAP =∠PAQ = 30°

• ∠APB = ∠ACB (Both angle made on same side of segment)

        So ∠APB = 60°

        Now we can write

        ∠APQ = 180 -∠APB  ( by linear pair angle)

        ∠APQ = 180 - 60

        ∠APQ = 120°

• From Δ APQ, use triangle angle property rules

       ∠PAQ +∠APQ +∠PQA = 180°

       30° +120° +∠PQA = 180°  

       On solving

       ∠PQA = 30° =∠BQA

• We have solved that ∠ACB = 60°,  So

       ∠BCQ = 180 -∠ACB ( by linear pair angle)

       ∠BCQ = 180 - 60

       ∠BCQ = 120°

• From Δ BCQ, use triangle angle property rules

       ∠BCQ +∠CQB +∠QBC = 180°

       120° +30° +∠QBC = 180°  

       On solving

      ∠QBC = 30°

• Now we can say that Δ BCQ is an isosceles triangle

        So

        BC = CQ  (side opposite to equal angle are equal)