Q6 of 31 Page 209

X is a point on the tangent at the point A lies on a circle with center O. A secant drawn from a point X intersects the circle at the points Y and z. If P is a mid-point of YZ, let us prove that XAPO or XAOP is a cyclic quadrilateral.

Formula used.


⇒ Perpendicular to tangent pass through centre of circle


⇒ Mid-point of chord is perpendicular line passes through centre


Solution.



As we join the figure


If P is mid-point of chord YZ


Then;


Line passing through centre to mid-point of line is perpendicular


Therefore OP is perpendicular to YZ


∠P = 90°


As there is tangent from point A on circle


Line passing through centre and point of contact is perpendicular to tangent


∠A = 90°


In Quadrilateral XAOP


∠A + ∠P = 90° + 90° = 180°


∠A + ∠P + ∠O + ∠X = 360°


∠O + ∠X = 360° - 180° = 180°


Sum of both opposite angles are 180°


∴ Quadrilateral XAOP is cyclic quadrilateral


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