Midpoints of diagonals

Question

Let $ABCD$ be convex quadrilateral such that $AB=CD$. And $E\neq F$ where $E, F$ is midpoint of $AC, BD$ respectively. Then prove that angle between$(AB, EF)$ and $(CD, EF)$ are equal.

I'll prove $\vec{AB}\cdot\vec{EF}=\vec{CD}\cdot\vec{EF}$. Using $\vec{EF}=\frac{1}{4}(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})$, it's sufficient to prove that $$(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{AB}=(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{CD}\iff (\vec{CB}+\vec{AD})\cdot\vec{AB}=(\vec{CB}+\vec{AD})\cdot\vec{CD}\iff BC\cdot CDcos\angle C+AD\cdot CDcos\angle D=BC\cdot ABcos\angle B+BA\cdot ADcos\angle A\iff BCcos\angle C+ADcos\angle D=CBcos\angle B+ADcos\angle A$$. I can't continue after this. Can anyone help me?

Answer 1

Let the position vectors of $A,B,C$ and $D$ be $a,b,c$ and $d$ (not writing arrow above the symbols for ease in writing). \begin{align} (\vec {AB}-\vec {CD})\cdot\vec{EF}&=(b-a-(d-c))\cdot\left(\dfrac{b+d}{2}-\dfrac{a+c}{2}\right)\\ &=\dfrac12\times(b-a+c-d)\cdot(b-a-(c-d))\\ &=\dfrac12\times(|b-a|^2-|c-d|^2)\\ &=\dfrac12\times(AB^2-CD^2)\\ &=0 \end{align}

Answer 2

Let $I$ be the midpoint of $AD$.
Then $|IE| = \frac{1}{2} |CD| = \frac{1}{2} |AB| = |IF|$.
So $IEF$ is an isosceles triangle.
It follows that $ \angle DHG = \angle IEG = \angle IFH = \angle AGH$.

Answer 3

It seems that my solution is to advanced, nobody want's to read it.

Translate $D$ for vector $\vec{CB}$ in to new point $G$. Then $BCDG$ is paralelogram, so $$\angle (EF,CD) = \angle (EF,BG)$$ Also, since diagonals in parallelogram halves each other we see that $F$ is on $CG$. But then $EF$ is midlline in triangle $ACG$ parallel to $AG$ and since triangle $ABG$ is isosceles we are done.