Question

In figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that $\Delta AEO\sim \Delta ABC$.

Answer 1

The figure given in the question is below

Let us first take up $△AOP$

We have,

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

Therefore $△AOP$ 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,

$\angle OEA$ = ${90}^{\circ }$

Now let us take up $△AOE$ and. $△ABC$

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,

$\angle ABC$ = ${90}^{\circ }$

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

$\angle OEA$ = ${90}^{\circ }$

Therefore,

$\angle ABC$ = $\angle OEA$

$\angle A$ is the common angle to both the triangles.

Therefore, from AA postulate of similar triangles,

$△AOE$ $\sim$ $△ABC$

Thus proved.