Prove that the tangent at the …
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Sia ? 4 years, 8 months ago
Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Let the two tangents AP and BP meets at P.
Now Join OP. Suppose OP meets AB at C and the chord AB=AC+BC.
We have to prove that {tex}\angle P A C = \angle P B C{/tex}
In two triangles PCA and PCB, we have
PA = PB [{tex} \because{/tex}Tangents from an external point are equal]
{tex}\angle A P C = \angle B P C{/tex} [{tex} \because {/tex} PA and PB are equally inclined to OP]
and, PC = PC [Common]
So, by SAS-criterion of congruence, we obtain
{tex}\Delta P A C \cong \Delta P B C{/tex}
{tex}\Rightarrow \quad \angle P A C = \angle P B C{/tex}
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