Question
Mon January 16, 2012 By:

Two Tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that angle APB = 2 angle OAB .

Mon January 16, 2012

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

angle OAP = angle OBP = 90Âº

Now,

angle OAP + angle APB + angle OBP + angle AOB = 360Âº [Angle sum property of quadrilaterals]

implies 90Âº+ angle APB + 90Âº + angle AOB = 360Âº

implies angle AOB = 360Âº - 180Âº - angle APB = 180Âº - angle APB ....(1)

Now, in triangle OAB, OA is equal to OB as both are radii.

implies angle OAB = angle OBA [In a triangle, angles opposite to equal sides are equal]

Now, on applying angle sum property of triangles in ?AOB, we obtain

Angle OAB + angle OBA + angle AOB = 180Âº

implies 2angle OAB + angle AOB = 180Âº

implies 2 angle OAB + (180Âº - angle APB) = 180Âº [Using (1)]

implies 2 angle OAB = angle APB

Thus, the given result is proved

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