[ Pythagorean Theorem (direct and inverse). ]  [ Properties and characteristics of an isosceles triangle. ]

In an isosceles triangle \(ABC\), the side \(AB\) is 10, and the base \(AC\) is 12. The angle bisectors of angles \(A\) and \(C\) intersect at point \(D\). Find \(BD\).
If $M$ is the midpoint of $A C$, then $B D: D M=B C: C M$.

## Solution

Since the angle bisectors of a triangle intersect at one point, $B D$ is the bisector of angle $B$. Extend $B D$ to intersect $A C$ at point $M$. Then $M$ is the midpoint of $A C, B M \perp A C$. Therefore, $B M^{2}=B C^{2}-M C^{2}=64$.

Since $C D$ is the bisector of triangle $B M C$, then $B D: D M=B C: C M=5: 3$.

Therefore, $B D=5 / 8 B M=5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c1015ea0b2d2805d8876g-02.jpg?height=363&width=529&top_left_y=2454&top_left_x=770)

Answer

## [ Properties and characteristics of isosceles triangles. ] $\left.\begin{array}{l}\text { Problem } 54303 \text { Topics: [Trigonometric ratios in a right triangle }] \\ \text { [ area of a triangle (using height and base). }]\end{array}\right]$

The base of an isosceles triangle is $b$, and the height dropped to the lateral side is $h$. Find the area of the triangle.

## Solution

Let $\alpha$ be the angle at the base, $H$ be the height dropped to the base of the triangle. Then

$$
\sin \alpha=\frac{h}{b}, \quad H=\frac{1}{2} b \operatorname{tg} \alpha=\frac{b h}{2 \sqrt{b^{2}-h^{2}}}, \quad S=\frac{1}{2} b h=\frac{b^{2} h}{4 \sqrt{b^{2}-h^{2}}}
$$

## Answer

$$
\frac{b^{2} h}{4 \sqrt{b^{2}-h^{2}}}
$$