# XY is a diameter

In the figure above, we have to prove that XA bisects angle YXZ.

Angle XAY = 90°

And XA=AY

Thus ∠AXY = AYX

In Δ AXY,

∠AXY+∠AYX+∠XAY =180° (As sum of angles of a triangle =180°

Thus,

2∠AXY=90° (∠AXY=∠AYX and ∠XAY=90°)

∠AXY=45°

∠ZXY=90°

Thus, ∠AXZ=45°

Hence, XA bisects ∠YXZ

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