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 .
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
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