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).

Since $$DM\perp AC$$ and $$M$$ is a midpoint of $$AC$$, we obtain $$AD=DC$$.
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)
• Yes. Thank you! $DM$ is the perpendicular bisector of $AC$. – Andrew Rogers Aug 19 '19 at 16:34
• Now I am trying to show that $HP \perp AB$ $(H = AC \cap BD)$. Is this true for all cases, because I can't prove it? – Andrew Rogers Aug 19 '19 at 16:52
• @Andrew Rogers Yes it's true because $HM=HN$ and $PM=PN.$ It's true if $AD=BC,$ which we got. – Michael Rozenberg Aug 19 '19 at 16:54
• I am not sure I understand why it's true iff $HM=HN$ and $PM=PN$. – Andrew Rogers Aug 19 '19 at 17:01