Let trapezoid $ABCD$ be a right trapezoid at $A$ and $D$. We set $AB=a$, $CD=b$, $AD=h$.

1) What relation must exist between $a, b, h$ for the diagonals $(AC)$ and $(BD)$ to be perpendicular?
2) Let $M$ be the midpoint of $[AD]$. What relation must exist between $a, b, h$ for the triangle $BMC$ to be right-angled at $M$?
1)

![](https://cdn.mathpix.com/cropped/2024_05_10_169e018874c7653739feg-33.jpg?height=371&width=572&top_left_y=2254&top_left_x=663)
$A D C$ and $B A D$ are similar because $\widehat{D A C}=90^{\circ}=\widehat{B A D}$, and $\widehat{D A C}=\widehat{C D B}$ because $(D A) \perp(D C)$ and $(A C) \perp(D B)$.

We deduce that $\frac{A D}{D C}=\frac{B A}{A D}$, so $\frac{h}{b}=\frac{a}{h}$, or $h^{2}=a b$.
2)

![](https://cdn.mathpix.com/cropped/2024_05_10_169e018874c7653739feg-34.jpg?height=455&width=764&top_left_y=607&top_left_x=732)

$A B M$ and $D M C$ are similar, so $\frac{A M}{A B}=\frac{D C}{D M}$, which can be written as $\frac{h}{2 a}=\frac{b}{h / 2}$, or $h^{2}=4 a b$.