\left.\begin{array}{l}{[\text { Theorem of the sum of squares of diagonals }]} \\ {[\text { Pythagorean Theorem (direct and inverse) }}\end{array}\right]

In parallelogram $A B C D$, the height drawn from vertex $B$ of the obtuse angle to side $D A$ divides it in the ratio 5:3, counting from vertex $D$. Find the ratio $A C: B D$, if $A D: A B=2$.
Apply the theorem on the sum of the squares of the diagonals of a parallelogram.

## Solution

Let $A D=8 x, A B=4 x$. Let $K$ be the base of the specified height. Then

$$
\begin{gathered}
A K=3 x, D K=5 x, B K=\sqrt{A B^{2}-K C^{2}}=x \sqrt{7} \\
B D^{2}=B K^{2}+K D^{2}=32 x^{2}
\end{gathered}
$$

Since

$$
A C^{2}+B D^{2}=2 \cdot A B^{2}+2 \cdot B C^{2}
$$

then

$$
A C^{2}+32 x^{2}=32 x^{2}+128 x^{2} .
$$

$$
A C^{2}=128 x^{2}, A C=8 x \sqrt{2}
$$

Therefore,

$$
\frac{A C}{B D}=\frac{8 x \sqrt{2}}{4 x \sqrt{2}}=2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-02.jpg?height=332&width=809&top_left_y=414&top_left_x=630)

## Answer

$2: 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-02.jpg?height=109&width=1212&top_left_y=943&top_left_x=0)

Prove that the ratio of the sum of the squares of the medians of a triangle to the sum of the squares of its sides is $\frac{3}{4}$.

## Hint

The square of the median of a triangle, drawn to side $a$, is equal to $\frac{1}{4}\left(2 b^{2}+2 c^{2}-a^{2}\right)$, where $a, b$ and $c$ are the sides of the triangle.

## Solution

Let the sides of the triangle be $a, b$, and $c$. If $m$ is the median drawn to side $a$, then

$$
m^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}-a^{2}\right)
$$

This can be proven by applying the theorem on the sum of the squares of the diagonals of a parallelogram (by preliminarily constructing the triangle into a parallelogram). Similarly for the squares of the other two medians. Adding the three obtained equations, we get the required result.