$3+$ [ Midline of a trapezoid

The diagonal of an isosceles trapezoid is equal to $a$, and the midline is equal to $b$. Find the height of the trapezoid.

#
The projection of the diagonal of an isosceles trapezoid onto the larger base is equal to the midline of the trapezoid.

## Solution

Let $H$ be the projection of vertex $C$ of the smaller base of trapezoid $A B C D$ onto the larger base $A D$. Then $C H$ is the height of the trapezoid, and the segment $A H$ is equal to the midline of trapezoid $A B C D$, since

$$
A H=A D-D H=A D-\frac{A D-B C}{2}=\frac{A D+B C}{2}
$$

From the right triangle $A C H$ we find that

$$
C H=\sqrt{A C^{2}-A H^{2}}=\sqrt{a^{2}-b^{2}} .
$$

Let $H$ be the projection of vertex $C$ of the smaller base of trapezoid $A B C D$ onto the larger base $A D$. Then $C H$ is the height of the trapezoid, and the segment $A H$ is equal to the midline of trapezoid $A B C D$, since

$$
A H=A D-D H=A D-\frac{A D-B C}{2}=\frac{A D+B C}{2}
$$

From the right triangle $A C H$ we find that

$$
C H=\sqrt{A C^{2}-A H^{2}}=\sqrt{a^{2}-b^{2}}
$$

Let $H$ be the projection of vertex $C$ of the smaller base of trapezoid $A B C D$ onto the larger base $A D$. Then $C H$ is the height of the trapezoid, and the segment $A H$ is equal to the midline of trapezoid $A B C D$, since

$$
A H=A D-D H=A D-\frac{A D-B C}{2}=\frac{A D+B C}{2}
$$

From the right triangle $A C H$ we find that

$$
C H=\sqrt{A C^{2}-A H^{2}}=\sqrt{a^{2}-b^{2}}
$$

## Answer

## Problem