Question

In Fig. BC is a tangent to the circle O. OE bisects AP. Prove that ΔAEO - ΔABC

Answer 1

Triangle AOP is an isosceles triangle because OA = OP as they are the radius of the circle. 

We know that radius of the circle is always perpendicular to the tangent at the point of contact.

Here 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 isoceles triangle we have found that

\(\angle ABC=90^\circ\)

Therefore,

\(\angle ABC=\angle OEA\)

\(\angle A\) is common angle to bothtriangle

Therefore, from AA postulates of similar triangle

ΔAOE \(\sim\) ΔABC