A circle with diameter the minor base $CD$ of a trapezium $ABCD$ intersects its diagonals $AC$ and $BD$ in, respectively, their midpoints $M$ and $N$. The lines $DM$ and $CN$ intersect in $P$ and $AC$ and $BD$ intersect in $H$. Show that $AD=CD=BC$ and $HP \perp AB$.
$DMNC$ is a cyclic quadrilateral and $CD||MN$, thus $DMNC$ is an isosceles trapezoid. Therefore, $CM=DN$ and $AC=BD$. Now I am trying to show $AD=CD$. $\angle DCM$ is inscribed and it's equal to $\angle DNM$ but I don't see how to compare it with $\angle CAD$. How is this done? For the second part of the problem, I tried to show that $HO$ passes through $P$ ($HO \perp CD$ because $\triangle CDH$ is isosceles and if we show $P \in HO$ we are done).
Best Answer
Since $DM\perp AC$ and $M$ is a midpoint of $AC$, we obtain $AD=DC$.
Can you end it now?
Since $DMNC$ is an isosceles trapezoid, we obtain: $$\measuredangle PMN=\measuredangle PDC=\measuredangle PCD=\measuredangle PNM,$$ which gives $$PM=PN.$$ Also, $$\Delta MDN\cong\Delta NCM,$$ which gives $$\measuredangle HMN=\measuredangle HNM,$$ which gives $$HM=HN,$$ which says that $PMHN$ is a kite, which says $$PH\perp MN.$$ About a kite see here: https://en.wikipedia.org/wiki/Kite_(geometry)