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.

In △ ABC,

CX is the angle bisector of ∠C.

⇒ ∠ACX = ∠BCX

⇒ ∠ACY = ∠BCX ……..(1)

In △ AXY,

⇒ AX = AY (Given)

⇒ ∠AXY = ∠AYX (Angles opposite to equal sides are equal) ………..(2)

From figure,

⇒ ∠XYC = ∠AXB = 180° (Since, XYC and AXB is a straight line)

⇒ ∠AYX + ∠AYC = ∠AXY + ∠BXY

⇒ ∠AXY + ∠AYC = ∠AXY + ∠BXY [From equation (2)]

⇒ ∠AYC = ∠AXY - ∠AXY + ∠BXY

⇒ ∠AYC = ∠BXY ……….(3)

By angle sum property of triangle AYC and BXC,

⇒ ∠AYC + ∠ACY + ∠CAY = ∠BXC + ∠BCX + ∠XBC

⇒ ∠BXY + ∠BCX + ∠CAY = ∠BXC + ∠BCX + ∠XBC [From equations (1) and (3)]

⇒ ∠BXC + ∠BCX + ∠CAY = ∠BXC + ∠BCX + ∠XBC [∵ From fig. ∠BXY = ∠BXC and ∠XBC = ∠ABC]

⇒ ∠CAY = ∠ABC.

Hence, proved that ∠CAY = ∠ABC.