## [ Auxiliary area. The area helps to solve the problem ]

 Pythagorean Theorem (direct and inverse)In triangle $A B C$, heights $A E$ and $C D$ are drawn. Find the side $A B$, if $B D=18, B C=30, A E=20$.
$A B \cdot C D=B C \cdot A E$

## Solution

By the Pythagorean theorem from the right triangle $B D C$, we find that

$$
C D^{2}=B C^{2}-B D^{2}=30^{2}-18^{2}=(30-18)(30+18)=12 \cdot 48=12^{2} \cdot 2^{2}
$$

thus, $C D=24$.

Let $S$ be the area of the triangle. Then

$$
S=\frac{1}{2} A B \cdot C D \text { and } S=\frac{1}{2} B C \cdot A E
$$

therefore, $A B \cdot C D=B C \cdot A E$, from which

$$
A B=\frac{B C \cdot A E}{C D}=\frac{30 \cdot 20}{24}=25
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-46.jpg?height=111&width=1875&top_left_y=242&top_left_x=0)

In the isosceles triangle $A B C$, the lateral sides $B C$ and $A C$ are twice the length of the base $A B$. The angle bisectors at the base intersect at point $M$. What fraction of the area of triangle $A B C$ is the area of triangle $A M B$?

## Hint

The angle bisector divides the base of the triangle into segments proportional to the lateral sides.

## Solution

Let $A A_{1}$ and $B B_{1}$ be the angle bisectors at the base $A B$ of triangle $A B C$. Since

$$
\frac{B A_{1}}{A_{1} C}=\frac{A B}{A C}=\frac{1}{2}, B A_{1}=\frac{1}{3} B C
$$

$$
S_{\triangle \mathrm{ABA}_{1}}=\frac{1}{3} S_{\Delta \mathrm{ABC}}
$$

Since $B M$ is the angle bisector of triangle $B A A_{1}$, then

$$
\frac{A_{1} M}{A M}=\frac{B A_{1}}{A B}=\frac{2}{3}, A M=\frac{3}{5} A A_{1}
$$

Therefore,

$$
S_{\Delta \mathrm{AMB}}=\frac{3}{5} S_{\Delta \mathrm{ABA}_{1}}=\frac{1}{5} S_{\Delta \mathrm{ABC}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-46.jpg?height=412&width=558&top_left_y=1592&top_left_x=750)

## Answer