Tangent PC intersects circle O at C, chord AB II CP, diameter COD intersects AB at E, and diameter AOF is extended to P.
Prove that $\Delta OPC \sim \Delta OAE$

Hint: We have to find the condition to make the given two triangles similar. Basic concepts about parallel lines and angles are required. We should also know the criteria to prove similarity between two triangles.

Complete step-by-step answer:
Given: PC is tangent to the circle with centre O, intersecting the circle at C. CD and AF are diameters.
And, AB II CP.
In the given figure,
$\angle AOE = \angle COP$
as they are vertically opposite angles.
Also it is given that AB II CP
\[\therefore \angle OAE = \angle CPO\]












This is because they are alternate interior angles.
Now, in $\Delta OPC\,and\,\Delta OAE$
$\angle AOE = \angle COP$
\[\angle OAE = \angle CPO\]
Therefore, we can say that
$\Delta OPC \sim \Delta OAE$ …by AA property of similarity of triangles.
Hence proved, the given two triangles are similar.

Note: Similarity of triangles is a very elementary concept and proving it very easy too. The other properties to prove the similarity between two triangles are SSS (Side Side Side) and SAS (Side Angle Side). AA property is also the same as AAA because if two angles of a given pair of triangles are equal, we can calculate the third angle which would also be equal.