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Question
ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX = AY.
Prove that: ∠CAY = ∠ABC.
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Solution

In ABC,
CX is the angle bisector of ∠C
⇒ ∠ACY = ∠BCX .........(i)
In ΔAXY,
AX = AY .........[Given]
∠AXY = ∠AYX ........(ii) [angles opposite to equal sides are equal]
Now,
∠XYC = ∠AXB = 180° .........[straight line]
⇒ ∠AYX + ∠AYC = ∠AXY + ∠BXY
⇒ ∠AYC = ∠BXY .......(iii) [From (ii)]
In ΔAYC and ΔBXC
∠AYC + ∠ACY + ∠CAY = ∠BXC + ∠BCX + ∠XBC = 180°
⇒ ∠CAY = ∠XBC .......[From (i) and (iii)]
⇒ ∠CAY = ∠ABC
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