4
$\begingroup$

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 $HP \perp AB$.

enter image description here

$DMNC$ 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).

$\endgroup$
4
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ Yes. Thank you! $DM$ is the perpendicular bisector of $AC$. $\endgroup$ – Andrew Rogers Aug 19 '19 at 16:34
  • $\begingroup$ @Andrew Rogers You are welcome! $\endgroup$ – Michael Rozenberg Aug 19 '19 at 16:35
  • $\begingroup$ 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? $\endgroup$ – Andrew Rogers Aug 19 '19 at 16:52
  • $\begingroup$ @Andrew Rogers Yes it's true because $HM=HN$ and $PM=PN.$ It's true if $AD=BC,$ which we got. $\endgroup$ – Michael Rozenberg Aug 19 '19 at 16:54
  • $\begingroup$ I am not sure I understand why it's true iff $HM=HN$ and $PM=PN$. $\endgroup$ – Andrew Rogers Aug 19 '19 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.