contest-matheuclidean-geometrygeometryplane-geometryquadrilateral
Let $ABCD$ be a quadrilateral. Let $E$ and $F$ be midpoint of $AB$ and $BC$. The lines $DE$ and $DF$ intersect $AC$ at $M$ and $N$ respectively. Suppose that $AM$ $=$ $MN$ $=$ $MC$. Prove that $ABCD$ is a parallelogram.
What I know is $EM$ are parallel to $BN$ and also $EM$ length is half of $BN$. In the same way with $NF$ and $BM$. I also tried Appolonius Theorem but failed. Hint please. Thank you very much!
Well, Geogebra says it is $\approx 77,34^{\circ}$, so good luck...
Actually, Ceva might really help:
$${\sin 80\over \sin 40}{\sin(70-x)\over \sin x}{\sin 20\over \sin90} = 1$$
After some manipulation we get $$\cot x = \tan 20+{2\over \cos 10}\implies x =... $$
Let $M$ be the common midpoint of $\overline{AC}$ and $\overline{BD}$. Then $M$ is also the midpoint of $\overline{PR}$.
Reflect everything through $M$. Lines $AB$ and $CD$ are swapped. Lines $AD$ and $BC$ are swapped, so lines $PQ$ and $RS$ are swapped. So points $Q$ and $S$ are swapped.
So $M$ is the common midpoint of $\overline{PR}$ and $\overline{QS}$, which implies $PQRS$ is a parallelogram.
Best Answer
It suffices show that $AC$ bisects $BD$. Let $X$ be the intersection between $AC$ and $BD$. Since $E$ and $F$ are midpoints of $AB$ and $BC$, $BX$ is bisected by $EF$. If $Y$ is the point of intersection between $BD$ and $EF$, then $|XY|=|BY|$, so that $|XY|=\dfrac{|BX|}{2}$.
Let $\ell:=|AC|$. Then, $|MN|=\dfrac{\ell}{3}$. Because $|EF|=\dfrac{|AC|}{2}=\dfrac{\ell}{2}$, we conclude that $\dfrac{|MN|}{|EF|}=\dfrac{2}{3}$. Since $MN\parallel EF$, we conclude that the line $\dfrac{|DX|}{|DY|}=\dfrac{2}{3}$, implying $$|XY|=|DX|-|DY|=\dfrac{|DX|}{2}\,.$$ That is, $|DX|=2\,|XY|=|BX|$. Thus, $AC$ bisects $BD$, as desired.