Let $ABCD$ be an isosceles trapezoid where $AB||CD$ and $AD=BC$, and let $O$ be the intersection of $AC$ and $BD$. Let $M, N,$ and $P$ be the midpoints of $AO, DO$, and $BC$ respectively. Given that $\angle AOB=60^\circ$, prove that $\triangle MNP$ is equilateral.

I (think) AOB and OCD are equilateral, and that $ABCD$ and therefore $BCNM$ is cyclic. I'm trying to prove that $P$ is the circumcentre of $BCN$ - here's my (slightly naïve?) approach.

Since $N$ is the midpoint of $PO$ and $\triangle DCO$ is equilateral, $ONC=90^\circ$. Then by Thales' Theorem, $BC$ is a diameter of $\triangle BCN$'s circumcircle. Thus $BP=PC=PN$ are radii, and since $CBMN$ is cyclic, the circumcircle of $\triangle BCN$ passes through $M$. Thus $PM$ is also a radius, so $PN=PM$. Also, $MN=\frac{AD}{2}$(similarity) $=\frac{BC}{2}=PB=PM=PN$, so $\triangle PNM$ is equilateral as desired.

Is this proof correct?* I literally just came up with it (as I was writing what I had tried in solving the question) and I'm not sure if it's right.

*If not, what have I done wrong?

I'd be grateful for any help I can receive.

  • $\begingroup$ It looks correct to me. You should only adjust a misprint: "Since $N$ is the midpoint of $DO$" and explain in more detail why $BCNM$ is cyclic. $\endgroup$ – Aretino Jul 8 '18 at 18:23
  • $\begingroup$ I might be misunderstanding the question because I drew it and $\triangle MNP$ looks very far from being equilateral: i.imgur.com/JMsZSo6.png $\endgroup$ – Lazy Lee Jul 9 '18 at 17:33
  • $\begingroup$ @LazyLee My bad - $O$ should be the intersection of $AC$ and $BD$ (i.e. the two diagonals). I'll correct this in my main post. $\endgroup$ – user574848 Jul 10 '18 at 4:59

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