[Linear element relationships of similar triangles] [Rectangles and squares. Properties and characteristics]

In a triangle with a base of 30 and a height of 10, an isosceles right triangle is inscribed such that its hypotenuse is parallel to the base of the given triangle, and the vertex of the right angle lies on this base. Find the hypotenuse.
Use the equality of the ratios of corresponding heights to the bases in similar triangles.

## Solution

Let the vertices $M$ and $N$ of the hypotenuse lie on the sides $A C$ and $B C$ of triangle $A B C$, respectively, and the vertex of the right angle $K$ lies on the base $A B$. Denote by $x$ the height of triangle $M N K$ drawn from vertex $K$. Then $M N = 2x$.

The heights of similar triangles $M C N$ and $A C B$, drawn from vertex $C$, are in the same ratio as the bases $M N$ and $A B$, that is, $(10 - x) / 10 = 2x / 30$. Therefore, $x = 6$, and $M N = 2x = 12$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ef0b80dacb6d98eff5ag-32.jpg?height=429&width=712&top_left_y=2358&top_left_x=679)

![](https://cdn.mathpix.com/cropped/2024_05_06_0ef0b80dacb6d98eff5ag-33.jpg?height=117&width=1392&top_left_y=153&top_left_x=1)

$A B C$ is the given triangle; $C D$ is the angle bisector of angle $C$; point $E$ lies on side $B C$, and $D E \parallel A C$. Find $D E$ if $B C = a$ and $A C = b$.

## Hint

The angle bisector of a triangle divides the opposite side into segments proportional to the other two sides.

## Solution

By the property of the angle bisector of a triangle, $B D / A B = a / (a + b)$.

From the similarity of triangles $D B E$ and $A B C$, it follows that $D E : A C = B D : A B$, hence $D E = ab / (a + b)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ef0b80dacb6d98eff5ag-33.jpg?height=383&width=684&top_left_y=783&top_left_x=687)

Answer

$ab / (a + b)$.