The two angles marked $y ^{\circ}$ are equal because they are in an isosceles triangle. For the same reason, the angles $z^{\circ}$are equal. Since an exterior angle of a triangle is the sum of the two interior and opposite angles, it follows that $a=2y$ and $b=2z$. Now $a^{\circ}+ b^{\circ} = 180^{\circ}$ since they are the base angles of a parallelogram. So $2y + 2z = 180$ giving $y+z=90$. But, from the angle sum of a triangle $x+y+z=180$; hence $x =90$.

*This problem is taken from the UKMT Mathematical Challenges.*