In Figure, `B C` is a tangent to the circle with centre `OdotO E` bisects `A Pdot` Prove that ` A E O ~ A B Cdot`

Explanation

We have,

`OA=OP `(Radius of the circle)

`/_OEA=90^0 `

`/_ABC=90^0` ( tangent to the circle is perpendicular to the radius through the point of contact )

`:. /_OEA=/_ABC` (Right angle)

`:. /_A `is the common angle to both triangle

`:. /_\AOE~= /_\ABC` (by ASA congruence)