Bisectors of angles of a triangle ABC meet its circumcircle at points X, Y and Z respectively. Prove that the angles of triangle XYZ are 90⁰-angle A/2, 90⁰- angle B/2, 90⁰-angleC/2

@Chayanikabiswas:Here is the answer to your question.

Given: A Δ ABC in which bisectors of angles meet the circle at X, Y and Z respectively.

To prove:

Proof: We know that angles in a same segment are equal.

∴ ∠BYX = ∠BAX

⇒ ∠BYX =

Also, ∠BYZ = ∠BCZ

⇒ ∠BYZ =

∴ ∠BYX + ∠BYZ =

⇒ ∠XYZ =

⇒ ∠XYZ =

⇒ ∠XYZ = 90° –

Now, AZ subtends angles ∠ ACZ and ∠ AXZ at points C and X in the same segment.

∴ ∠ACZ  = ∠AXZ  

⇒ ∠AXZ =

Also, ∠AXY = ∠ABY

⇒ ∠AXY =

∴ ∠AXZ + ∠AXY =  

⇒ ∠ZXY =

⇒ ∠ZXY =

⇒ ∠ZXY =90° –

Similarly, we can prove that ∠ XZY = 90° –