$\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}$ 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? poor paint image


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 $PM\cong QN$.

  • $\begingroup$ I can't guess how the two couples of triangles $OTP,\ OTQ$ and $OMP,\ OMQ$ are related. Could it be a typo? Either case, this use of the SSA criterion is not what I'm looking for since this is for a secondary schooler $\endgroup$ – Eugenio Nov 20 '16 at 12:01
  • $\begingroup$ The congruence of triangles $OTP$ and $OTQ$ allows one to establish that $OP\cong OQ$ and $\angle OPT\cong\angle OQT$. If SSA is not allowed, the same result could be obtained by the sine law applied to triangles $OMP$ and $ONQ$. $\endgroup$ – Intelligenti pauca Nov 20 '16 at 12:48
  • $\begingroup$ You're right, my bad. Not for the sine law though, it's not allowed as well. I edited the question ti make it clearer $\endgroup$ – Eugenio Nov 20 '16 at 12:55
  • $\begingroup$ I edited my answer to add a proof not relying on SSA. $\endgroup$ – Intelligenti pauca Nov 20 '16 at 14:21

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