Let $ABCD$ be a cyclic quadrilateral with diagonals intersecting at $T$. Let $P$ and $Q$ be the projections of $T$ onto $AB$ and $CD$ respectively. Let $X$ and $Y$ be the mid-points of $AD$ and $BC$ respectively.
I am required to prove that $P$ and $Q$ are symmetric in the line $XY$, but I have absolutely no idea where to begin here. All that comes to mind is that I could try to prove that the midpoint of $PQ$ lies on the line $XY$, but this does not seem like it would lead to anything.
Let $Z$ be a point such that $\angle DAZ = \angle ZDA = \angle PAT$ and $Z$, $T$ lie on different sides of the line $AD$. Then $\triangle PAT \sim \triangle XAZ$, therefore $$\frac{PA}{XA} = \frac{TA}{ZA}.$$ Moreover $$\angle PAX = \angle PAT + \angle TAX = \angle XAZ + \angle TAX = \angle TAZ.$$ Using SAS we conclude that $\triangle PAX \sim \triangle TAZ$. In particular $$\frac{XP}{ZT} = \frac{PA}{TA}.$$ Analogously $$\frac{XQ}{ZT}=\frac{QD}{TD}.$$ However $\triangle PAT \sim \triangle QDT$ so $$\frac{PA}{TA}=\frac{QD}{TD}.$$ Combining these three equalities we get $$\frac{XP}{ZT}=\frac{PA}{TA}=\frac{QD}{TD}=\frac{XQ}{ZT},$$ therefore $XP=XQ$.
We can prove in a similar way that $YP=YQ$. Therefore $P,Q$ are symmetric with respect to $XY$.