[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part Rectangles and squares. Properties and criteria

In a square $A B C D$ with side $a$, a circle is inscribed, touching side $C D$ at point $E$.

Find the chord connecting the points where the circle intersects the line $A E$.

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Let $P E$ be the desired chord, $M$ be the point of tangency of the circle with side $A D$ (see figure). Then $A M=\frac{a}{2}, A E^{2}=$ $A D^{2}+D E^{2}=5 a^{2} / 4$

By the tangent-secant theorem $A P=\frac{A M^{2}}{A E}=\frac{a}{2 \sqrt{5}}, \quad P E=A E-A P=\frac{2 a}{\sqrt{5}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4fce0705b02fa9e106d4g-09.jpg?height=383&width=328&top_left_y=2241&top_left_x=865)

## Answer

The bisector of the external angle at vertex $C$ of triangle $A B C$ intersects the circumcircle at point $D$.

Prove that $A D=B D$.

## Solution

Let $\angle C=2 \alpha$. Then $\angle A B D=\angle A C D=1 / 2\left(180^{\circ}-2 \alpha\right)=90^{\circ}-\alpha$. But $\angle B A D=180^{\circ}-2 \alpha-\left(90^{\circ}-\alpha\right)=90^{\circ}-\alpha$, so triangle $A D B$ is isosceles.

![](https://cdn.mathpix.com/cropped/2024_05_06_4fce0705b02fa9e106d4g-10.jpg?height=737&width=717&top_left_y=429&top_left_x=682)

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