a) Let $A$, $B$, $C$, $D$ be the vertices of a quadrilateral and $A'$, $B'$ respectively the projection of points $A$, $B$ on the opposite sides. Verify that when the lines $AA'$ and $BB'$ cut, then $AB\perp CD$.
b) Is the Reverse Statement True?
Please, can you solve this in two ways?
(1) Let us assume that the lines $AA'$, $BB'$ intersect in a point $H$.
So all points $A,A', B, B', H$ are points of the same plane $(\pi)$, say, as a matter of notation.
Also, from $AA'\perp (BCD)$ follows that $AA'$ is perpendicular on any line of the plane $(BCD)$, in particular $AA'\perp CD$.
Also, from $BB'\perp (ACD)$ follows that $BB'$ is perpendicular on any line of the plane $(ACD)$, in particular $BB'\perp CD$.
Two different lines of the plane $(\pi)$ are perpendicular to $CD$, so $(\pi)\perp CD$. The line $AB$ is also a line of $(\pi)$, so $AB\perp CD$.
$\blacksquare$
Note: An important role is played by the point $X$ of intersection of $CD$ and $(\pi)$.
(2) Assume $AB\perp CD$. Let $X$ be the projection of $A$ on $CD$ in the triangle $ACD$. From $AX\perp CD$ and $AB\perp CD$ it follows that the plane $(P)=ABX$ is perpendicular on $CD$. In particular, $BX\perp CD$.
Let now $A''$ be the projection of $A$ on $BX$ in the triangle $AXB$.
Then $AA''\perp BX$, and $AA''\perp CD$ (since it is a line in $(P)$), so $AA''$ is perpendicular on two different lines of the plane $(BCD)$, so $AA''\perp (BCD)$. This implies $A''=A'$, is the projection of $A$ on $(BCD)$.
Similarly, the projection of $B$ on $(ACD)$ also lies in the plane $(P)$.
So $AA'$ and $BB'$ intersect each other, being non-parallel lines in the same plane $(P)$.
$\blacksquare$