Acute angle $A$ of rhombus $A B C D$ is $45^{\circ}$, the projection of side $A B$ on side $A D$ is 12. Find the distance from the center of the rhombus to side $C D$.
Use the theorem of the midline of a triangle.

## Solution

Let $M$ be the projection of vertex $B$ of the rhombus $A B C D$ onto side $A D$. According to the problem, $A M=12$. The acute angle at vertex $A$ of the right triangle $A M B$ is $45^{\circ}$, so $B M=A M$.

Let $P$ and $Q$ be the projections of the center $O$ of the rhombus $A B C D$ onto sides $A D$ and $C D$ respectively. From the equality of right triangles $O D P$ and $O D Q$ (by hypotenuse and acute angle), it follows that $O Q=O P$, and since $O P$ is the midline of triangle $A B M$, then

$$
O P=\frac{1}{2} B M=\frac{1}{2} A M=6
$$

## Answer

6.