[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part [Pythagorean Theorem (direct and inverse).

A circle is tangent to side $B C$ of triangle $A B C$ at its midpoint $M$, passes through point $A$, and intersects segments $A B$ and $A C$ at points $D$ and $E$ respectively. Find the angle $A$, if it is known that $B C=12, A D=3.5$ and $E C=\frac{9}{\sqrt{5}}$.

#
Apply the tangent and secant theorem.

## Solution

By the tangent and secant theorem, \( B M^{2}=A B \cdot B D, C M^{2}=A C \cdot C E \), or \( 36=A B\left(A B-7 / 2\right), 36=\frac{9 A C}{\sqrt{5}} \), from which \( A B=8, A C=4 \sqrt{5} \).

Since \( A B^{2}+A C^{2}=64+80=144=B C^{2} \), triangle \( A B C \) is a right triangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-28.jpg?height=460&width=788&top_left_y=2120&top_left_x=635)

## Answer

\( 90^{\circ} \).

In the right triangle \( A B C \), a square is inscribed such that two of its vertices lie on the hypotenuse \( A B \), and the other two - on the legs. The radius of the circle circumscribed around triangle \( A B C \) is in the ratio 13:6 to the side of the square. Find the angles of the triangle.

## Hint

Express the hypotenuse of the triangle in terms of the side of the square and the tangent of the acute angle of the triangle.

## Solution

Let the vertices \( M \) and \( N \) of the square \( M N K L \) lie on the hypotenuse \( A B \) ( \( M \) - between \( N \) and \( B \) ). Denote the side of the square by \( x \), \( L A = \alpha \), and the radius of the circumscribed circle around triangle \( A B C \) by \( R \).

From the right triangles \( A N K \) and \( B M L \), we find that

\[
A N=\frac{x}{\operatorname{tg} \alpha}, B M=x \operatorname{tg} \alpha
\]

Therefore,

\[
2 R=A B=A N+N M+M B=\frac{x}{\operatorname{tg} \alpha}+x+x \operatorname{tg} \alpha
\]

Then,

\[
\frac{1}{\operatorname{tg} \alpha}+\operatorname{tg} \alpha+1=\frac{2 R}{x}=\frac{13}{3}
\]

or

\[
3 \operatorname{tg}^{2} \alpha-10 \operatorname{tg} \alpha+3=0
\]

From this equation, we find that \( \operatorname{tg} \alpha=3 \) or \( \operatorname{tg} \alpha=\frac{1}{3} \).

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-29.jpg?height=417&width=924&top_left_y=1555&top_left_x=567)

## Answer

\( \operatorname{arctg} 3, \operatorname{arctg} \frac{1}{3} \)