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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 .
Asked by | 16 Jan, 2012, 11:11: AM

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 = 90º

Now,

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

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

implies angle AOB = 360º - 180º - angle APB = 180º - 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 = 180º

implies 2angle OAB + angle AOB = 180º

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

implies 2 angle OAB = angle APB

Thus, the given result is proved

Answered by | 16 Jan, 2012, 02:51: PM

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