# Let the opposite sides of a cyclic quadrangle be extended to meet at V and W. Then the internal bisectors of the angles at V and W are perpendicular.

I have been trying to solve this problem and am stumped. Diagram below. My approach has been trying to show that $\Delta VKM$ is isosceles, since the bisector of the unequal angle of an isosceles triangle is perpendicular to the opposite side (which in this case is the bisector at W). I have gone down various blind alleys attempting to show that $\angle LKV = \angle LMV,$ many of them involving a line l through V and parallel to t (not shown, since I think this is the wrong approach). Although this gets me annoyingly close to my goal, I always wind up subtly assuming something that implies what I am trying to prove (for example that a perpendicular to l, t is parallel to s). The hint just says to compare the arcs into which the circle is divided by the bisectors at V and W, but I am not seeing it.

This is problem 3.2.8 in Coxeter and Greitzer's Geometry Revisited. Thanks for any help, I would really appreciate it.

How about direct simple angle chasing? Let the angle at $A$ be $\alpha$ and the angle at $B$ be $\beta$. Then the angle at $V$ is $180^\circ-\alpha-\beta$ and the angle at $W$ is $\beta - \alpha$.
We can calculate that $\angle LWC = \frac{ \beta-\alpha}{2}$, $\angle WCV = 180^\circ+\alpha$ (taking the interior angle of the quadrilateral), $\angle CVL = \frac{180^\circ-\alpha-\beta}{2}$. Hence,
$$\angle VLW = 360^\circ - (\frac{ \beta-\alpha}{2}) - (180^\circ+\alpha) - (\frac{180^\circ-\alpha-\beta}{2}) = 90^\circ.$$
Since the quadrilateral $ABCD$ is cyclic you may assume (as in your figure) that the angle at $A$ is ${\pi\over2}-\alpha$, at $B$ is ${\pi\over2}-\beta$, at $C$ is ${\pi\over2}+\alpha$, and at $D$ is ${\pi\over2}+\beta$ with $\alpha>\beta>0$. Then the angle at $V$ is $\alpha+\beta$ and at $W$ is $\alpha-\beta$. It is then easy to verify that the angle bisectors at $V$ and $W$ are orthogonal.