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.

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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.

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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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