Simple geometry problem - proving diagonals of quadrilateral are equal Let there be a convex quadrilateral ABCD. The midpoints of AD and BC are accordingly M and N. The angle of the intersection of diagonal AC and line MN is equal to the angle of the intersection of diagonal BD and the same line MN. Now, how does one prove that the diagonals are equal?
I've mingled with this here and there, trying different methods, but to no avail. I'll be thankful for any help!
Also, I apologise for any mistakes - English is not my first language.
 A: Let point of intersection of diagonals be O, intersection of AC and MN point P and intersection of BD and MN point R. 
Denote $\angle APM= \angle BRN =\angle CRN = \angle DRM = \theta $
Also$ \angle AMP= \phi$ then $\angle DMR = 180 - \phi $
and $ \angle BNR = \delta$ and $ \angle PNC = 180 - \delta
$
sine law in triangles APM and DMR gives us ; 
$\frac{AP}{AM}= \frac{\sin \phi}{\sin \theta} \Rightarrow AP= AM \cdot \frac{\sin \phi}{\sin \theta} $
$ \frac{RD}{MD}= \frac{\sin(180-\phi)}{\sin \angle DRM}= \frac{\sin \phi}{\sin \theta} \Rightarrow RD= MD \cdot \frac{\sin \phi}{\sin \theta} $
Since $AM=MD$ it follows that $AP=RD$. Analoguosly you can find that $PC=BR$ and from there 
$AP+PC=BR+RD \rightarrow AC=BD$
A: 
Let P, Q be the midpoints of AB and CD respectively. Then, … (by midpoint theorem) … , we have
(1) AC = 2PN and BD = 2QN; AND
(2) MPNQ is a parallelogram.
For some reasons (like alternate angles or corresponding angles), x = y ..... [3]
(2) and [3] imply MPNQ is a rhombus.
Result follows by applying (1).
