Problem 10.7. Two circles have a common chord $AB$. Through point $B$, a line is drawn that intersects the circles at points $C$ and $D$, such that $B$ lies between $C$ and $D$. The tangents to the circles, drawn through points $C$ and $D$, intersect at a point $E$. Compare $AD \cdot AC$ with $AB \cdot AE$.
## Solution.

We apply the property of the angle measure with the vertex on the circle, one side tangent and the other secant to the circle, and the measure of the inscribed angle in the circle.

Let $m(\angle B A D)=\alpha \Rightarrow m(\operatorname{arc} B D)=2 \alpha$, then $m(\angle C D E)=\alpha$.

Let $m(\angle B A C)=\beta \Rightarrow m(\angle D C E)=\beta$.

From triangle $C D E$ we have $m(\angle C E D)=180^{\circ}-\alpha-\beta$. We observe that $m(\angle C A D)+m(C E D)=180^{\circ}$, so quadrilateral $A C E D$ is cyclic,

thus $\angle B D A \equiv \angle C E A$. (*) From $\left(^{*}\right)$ and the fact that $m(\angle B A D)=m(\angle C A E)=\alpha$,

![](https://cdn.mathpix.com/cropped/2024_06_06_b6f59016f22a8870cc7cg-1.jpg?height=596&width=851&top_left_y=1992&top_left_x=1088)
it follows that triangles $B A D$ and $C A E$ are similar, then $\frac{A B}{A C}=\frac{A D}{A E} \Rightarrow A D \cdot A C=A B \cdot A E$.