$\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}$ Let $AB$ be a chord on a circle and let $T$ be its midpoint. Choose two points $P$ and $Q$ on the chord such that $\overline{QT} = \overline{PT}$.
Lead two orthogonal lines from $P$ and $Q$ and let $M,N$ be the respective intersection of the lines with the arc $\arc{AB}$.
Prove that $MP\cong NQ$.
My attempt: I extended the orthogonal lines and defined $R,S$ so that I could apply the theorem on angles and parallel lines i.e. I obtained that the two couples of triangles are congruent: $MPT\cong SQT$ and $PRT \cong NQT$.
My idea is to prove that for example $MT\cong RT$ or $\widehat{TMP} = \widehat{TRP}$ so that the triangles $MPT,\ PRT$ are congruent.
EDIT This exercise is meant to be for a high schooler; therefore the answer should use simple arguments base on angles and triangles congruences and so on.
Any hints to do that?
It is easy to prove that triangles $OTP$ and $OTQ$ are congruent. Compare then triangles $OMP$ and $ONQ$: they have two couples of congruent sides and $\angle OPM\cong\angle OQN$. They are congruent because SSA criterion works if the sides opposite to congruent angles are greater than the other couple of sides, which is indeed our case.
EDIT: Proof without using SSA.
From $O$ draw a line parallel to $AB$, intersecting $PM$ and $QN$ at $H$ and $K$ respectively. $HKQP$ is a rectangle by construction, so we have $PH\cong QK$.
Consider now right triangles $OHM$ and $OKN$: they are congruent by hypotenuse-leg criterion (notice that $O$ is the midpoint of $HK$), so we have $HM\cong KN$.
By subtracting (or adding, depending on the position of line $AB$) those two equations one then gets $PM\cong QN$.