XY is a diameter of a circle. PAQ is a tangent to the circle at the point. ‘A’ lying on the circumference. The perpendicular drawn on the tangent of the circle from X intersects PAQ at Z. Let us prove that XA is a bisector of ∠YXZ.

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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