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In the given figure, *BC* is a tangent to the circle with centre *O*. *OE* bisects *AP.* Prove that ΔAEO ∼ Δ ABC.

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#### Solution

The figure given in the question is below

Let us first take up Δ AOP.

We have,

*OA = OP* (Since they are the radii of the same circle)

Therefore, Δ AOP. is an isosceles triangle. From the property of isosceles triangle, we know that, when a median drawn to the unequal side of the triangle will be perpendicular to the unequal side. Therefore,

`∠ OEA=90^o`

Now let us take up Δ AOE and Δ ABC..

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. In this problem, *OB* is the radius and *BC* is the tangent and *B* is the point of contact. Therefore,

`∠ ABC=90^o`

Also, from the property of isosceles triangle we have found that

`∠ OEA = 90^o`

Therefore,

`∠ ABC`= `∠ OEA`

`∠ A` is the common angle to both the triangles.

Therefore, from AA postulate of similar triangles,

`ΔAOE` ~ `ΔABC`

Thus we have proved.

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