# Two circles intersect at points A and B. A common tangent touches the first circle at point C and the second at point D. Let B be inside the triangle ACD. Let the line CB intersect the second circle again at point E. Prove that AD bisects the angle CAE.

Hello!

This problem requires one relatively simple fact:

Consider a tangent line to a circle and a secant line through the point of tangency. Then the measure of arc between the secant and the tangent is twice as much as the angle between a secant and a tangent.

You can...

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Hello!

This problem requires one relatively simple fact:

Consider a tangent line to a circle and a secant line through the point of tangency. Then the measure of arc between the secant and the tangent is twice as much as the angle between a secant and a tangent.

You can look at the simple proof with the link supplied.

Now to the problem. Look at the picture.

Denote the angle BCD as `alpha` and the angle BDC as `beta`. By the above statement the angle CAB is also `alpha` (it is based on the arc CB and also a half of it). Also, the angle DAB is `beta` (it is based on the arc BD of the second circle).

Next, tha angle DBE is `alpha+beta` as an external angle of a triangle CBD. And the angle DAE is equal to the angle DBE because they have the same base DE in the second circle.

So both CAD and DAE are `alpha+beta,` QED.

(many thanks to my old math teacher N.N.)

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