Prove that one diagonal of a quadrilateral bisects the other I tutor GCSE maths and I am stumped by this question:


*

*Let XBYA be a quadrilateral

*The diagonals AB and XY intersect at point M 

*Given that the area of triangle AXB = the area of triangle AYB


Prove that XY is bisected by AB.
(I am guessing that students are expected to use the area of a triangle = 1/2 ab sin(C) but I cannot show XM=MY? )
 A: Hints: 


*

*Draw perpendicular segments from $AB$ to $X$ and $AB$ to $Y$.  These are the altitudes of the two given triangles.

*Since the triangles share the base $AB$, to be the same area, they must have the same height (given by these altitudes).

*The triangles formed by these altitudes and $M$ are similar right triangles with one leg the same length, therefore, they are congruent.

*Their hypotenuses are $XM$ and $MY$, so $XM$ and $MY$ are the same length.
A: Let $P$ and $Q$ be the orthogonal projections of points $X$ and $Y$ respectively onto the diagonal $AB$. Then the segment $XP$ is the altitude of triangle $ABX$ from vertex $X$ to $AB$ and analogously,the segment $YQ$ is the altitude of triangle $ABY$ from vertex $Y$ to $AB$. Consequently
$$\frac{1}{2} \, XP \cdot AB =  \text{Area}(ABX) =  \text{Area}(ABY) = \frac{1}{2} \, YQ \cdot AB$$ so after you cancel out the common factors on both sides of the latter identity you get
$$XP = YQ$$
Moreover, $\angle \, XPM = 90^{\circ}=\angle \, YQM$ and $\angle \, XMP = \angle \, YMQ$ so triangles $MPX$ and $MQY$ are congruent, which yields $MX = MY$. 
