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.
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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