## [ Properties and characteristics of an isosceles triangle. Pythagorean theorem (direct and inverse). $\quad]$ Inscribed angle subtending the diameter $\quad]$

On the lateral side of an isosceles triangle, a circle is constructed with this side as the diameter, dividing the second lateral side into segments of lengths $a$ and $b$.

Find the base of the triangle.
Let the circle $\Omega$, constructed with the side $AB$ of the isosceles triangle $ABC$ as its diameter, intersect the side $AC$ at point $K$, such that $CK = a$ and $AK = b$.

First method. $\angle AKB = 90^\circ$. By the Pythagorean theorem, $BK^2 = AB^2 - AK^2 = (a+b)^2 - b^2 = a(a+2b)$, and $BC^2 = BK^2 - CK^2 = a(a+2b) + a^2 = 2a(a+b)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d3cd76ad0dc53ae543bdg-33.jpg?height=377&width=374&top_left_y=2058&top_left_x=842)

Second method. Let $\Omega$ intersect the base $BC$ at point $M$. Then $\angle AMB = 90^\circ$, meaning $AM$ is the height of the isosceles triangle, drawn to the base. Therefore, $M$ is the midpoint of $BC$.

From point $C$, two secants are drawn to the circle, so $CM \cdot CB = CK \cdot CA$, or $\frac{1}{2}BC^2 = a(a+b)$.

## Answer

$\sqrt{2a(a+b)}$ or $\sqrt{2b(a+b)}$.

## Pythagorean Theorem (direct and inverse). ] [Parallel lines, properties and criteria. Secants]

$AB$ and $CD$ are parallel lines, $AC$ is a secant (points $B$ and $D$ are on the same side of line $AC$), $E$ and $F$ are the points of intersection of lines $AB$ and $CD$ with the angle bisectors of angles $C$ and $A$. It is known that $AF = 96$ and $CE = 110$. Find $AC$.

## Solution

Let the segments $AF$ and $CE$ intersect at point $O$. Triangle $AOC$ is clearly a right triangle, with $AO = OF = 48$ and $CO = OE = 55$. Therefore,

$AC^2 = AO^2 + CO^2 = 48^2 + 55^2 = 73^2$

## Answer

73.