An isosceles trapezoid has parallel sides of lengths $a$ and $b$, and the projection of the midpoint of one leg onto the other leg falls at one of the endpoints of this leg. Calculate the area of the trapezoid.
Note: An isosceles trapezoid is typically understood to be symmetric with respect to the perpendicular bisector of its parallel sides, and this is what we will mean by it in the following. According to the meaning of the term, every parallelogram is also an isosceles trapezoid (and has equal parallel sides). Thus, if $a=b$, and we also consider parallelograms, then drawing any parallelogram from any point on the semicircle erected over an $a$-length $EF$ segment as a diameter, with $EF$ as its midline, we get a "isosceles trapezoid" that meets the conditions, so the area is indeterminate. Thus, we only exclude trivial cases. We can also exclude the case $a=b$, because in this case, according to the accepted interpretation, our trapezoid is a rectangle, and thus the projection of the midpoint of each side onto the opposite side is the midpoint of that side.

![](https://cdn.mathpix.com/cropped/2024_05_02_c65da787fd9aa6686cafg-1.jpg?height=452&width=590&top_left_y=506&top_left_x=754)

I. Solution. If $a$ and $b$ are known, it is sufficient to determine the height to calculate the area. For this, we first need to understand the shape of the trapezoid. Let the midpoint of the $DA$ leg be $E$, and its projection on the $BC$ leg be the point $C$. Then $E$ is in the interior of the angle domain of $\triangle BCD$, and thus

$$
\angle DCB > \angle ECB = 90^\circ
$$

so there is an obtuse angle at $C$ and, by symmetry, also at $D$. Thus, if $a > b$, then $AB = a$ and $CD = b$.

Reflect the $\triangle CDE$ over point $E$. Then point $D$ will map to point $A$, and point $C$ will map to the intersection point $C'$ of the extensions of segments $CE$ and $BA$. Thus, the area $t$ of the trapezoid is equal to the area of the right triangle $\triangle BCC'$. Let the projection of point $C$ onto $AB$ be $M$ and $CM$ be $m$, so

$$
t = \frac{BC' \cdot m}{2}
$$

and, using the geometric mean property of the height in a right triangle,

$$
m^2 = BM \cdot C'M, \quad \text{where} \quad BC' = a + b
$$

To determine $BM$, consider the projection of point $D$ onto $AB$ as point $N$. On one hand, $MN = b$, and on the other hand, by symmetry, $AN = BM$ and since the two together give the difference of the parallel sides, thus

$$
\begin{gathered}
BM = \frac{a - b}{2}, \quad C'M = a + b - \frac{a - b}{2} = \frac{a + 3b}{2} \\
m = \frac{\sqrt{(a - b)(a + 3b)}}{2}
\end{gathered}
$$

and the area of the trapezoid is

$$
t = \frac{(a + b) \sqrt{(a - b)(a + 3b)}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_c65da787fd9aa6686cafg-1.jpg?height=407&width=1047&top_left_y=2198&top_left_x=520)

Figure 2

II. Solution. To determine the height $m$, note that due to the symmetry of the trapezoid, the projection of the midpoint $F$ of the $BC$ leg onto the $AD$ leg is the point $D$, and the right triangles $\triangle CEF$ and $\triangle DFE$ are congruent. Reflect the $\triangle CEF$ over $EF$. Since the midline is equidistant from $AB$ and $CD$ and is parallel to them, the reflection of $C$ is the projection of $M$ on $AB$. The quadrilateral $DEMF$ has opposite sides equal, so it is a parallelogram, and it is a rectangle because the angles at $D$ and $M$ are right angles. Therefore, its diagonals are equal: $DM = EF = (a + b) / 2$, and from the right triangle $\triangle CDM$,

$$
m^2 = CM^2 = DM^2 - CD^2 = \left(\frac{a + b}{2}\right)^2 - b^2 = \frac{(a + 3b)(a - b)}{4}
$$

Note: There are several other ways to determine $m$, or we can calculate the length of the $BC$ leg and then the length of the leg $CC'$ of the triangle $\triangle BCC'$ mentioned in the first solution, and the area can be calculated from the formula $t = BC \cdot CC' / 2$. The above two calculations appear to be the simplest.

![](https://cdn.mathpix.com/cropped/2024_05_02_c65da787fd9aa6686cafg-2.jpg?height=398&width=613&top_left_y=736&top_left_x=737)