Leg bisectors in a trapezoid and angle equality

Question

Let $PQRS$ be a trapezoid. The bisectors of legs $PS$ and $QR$ intersect the opposite legs at $M$ and $N$ so that triangles $PMS$ and $QNR$ are formed.

Prove that $\measuredangle PMS= \measuredangle QNR$.

The question may seem easy but I fail to move on, playing with angles didn't help. I'd appreciate some hint from somebody with an eye for geometry and triangles.

Answer 1

Hint:

You have to show that triangles $PMS$ and $QNR$ are isosceles and angles $PSM,SPM,NRQ,NQR$ are congruent

$HK\parallel PQ\parallel SR$

Answer 2

If you extend the trapzoid legs they will eventually intersect (unless it's a parallellogram - for which the result should be rather obvious by symmetry) at a point $A$. The triangles $AKN$ and $AHM$ are similar since they have two pairs of congruent angles.

Without loss of generality we can assume that it happens above the trapezoid.

Also since $SR$ and $HK$ are parallel $AS$ it to $SH$ like $AR$ to $RK$. Which will mean that the triangles $RKN$ and $SHM$ are similar.

For thecase where it's a parallellogram we have that the angles at $K$ and $H$ are right and since $SP$ and $RQ$ are equal and $H$ and $K$ are the midpontsso $HM$ and $KR$ are equal an the $HMNK$ is a rectangle so $HM$ and $NK$ are equal so the triangles $KNR$ and $HMP$ are congruent and therefore similar.