Q. 24.5( 4 Votes )

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.

Answer :


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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