Question

O is the centre of the circle ,AB is a diameter OA=AP, O-A-P is a tangent through C. A tangent through point A intersects PC in E and BC in D. To prove. Triangle CED is an equilateral triangle

Answer 1

Answer:

Step-by-step explanation:

In the right triangle OCP.

Let r be the radius

OC = r

OP = OA + AP = 2r

Then PC is equal to the square root of OP square minus OC square, which will be equal to square root of 3r.

Hence, the ratio of the sides of triangle OCP is equal to the quotient between 2 and square root of 3.

Thus, its angles should be 30º, 90º and 60º.

Angle OPC and angle APE are the same, equal to 30º.

Angle APE is equal to 30º and angle AEP is equal to 60º.

Then angle CED is equal to 60º, since they are vertically opposite angles.

Then OB = OC, which means that angle OBC and angle OBC are both equal to 30º.

Hence proven.