In the figure, $ B C $ is a tangent to the circle with centre $ O $. OE bisects $ A P $. Prove that $ \triangle A E O \sim \triangle A B C $.
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Given:

In the figure, \( B C \) is a tangent to the circle with centre \( O \). OE bisects \( A P \).

To do:
We have to prove that \( \triangle A E O \sim \triangle A B C \).

 Solution:

In $\triangle OAE$ and $\triangle OPE$,

$OE = OE$  (Common side)

$OA = OP^{\prime}$ (radii of the same circle)

$EA = EP$     (Given)

Therefore, by SSS axiom,

$\triangle OAE = \triangle OPE$

$\angle OEA = \angle OEP$

$\angle OEA + \angle OEP = 180^o$

$\angle OEA = 90^o$

In $\triangle AEO$ and $\triangle ABC$,

$\angle OEA = \angle ABC = 90^o$

$\angle A = \angle A$   (Common angle)

Therefore, by AA axiom,

$\triangle AEO\ \sim\ \triangle ABC$

Hence proved.

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Updated on: 10-Oct-2022

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