AB is a diameter of a circle. BC is the tangent at B as shown in the given figure. Show that PBC = BAP.
ABC=90o Since AB being diameter is perpendicular to tangent BC at the point of contact.
So ABP +PBC =90o (i)
Also APB =90o (angle in the semi-circle)
So BAP+ABP = 90o (ii) (using angle sum property of triangles)
From (i) and (ii),
PBC = BAP
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