<Qua-2024-727>, hint

Let P, Q, R, S be the intersections of bisectors.

Then <RPS=<RQS=90, hence P,Q,R,S lie on a circle.

Suppose perpendiculars at A and B meet at R'. Since RPSQ and RBR'A are similar

BA and PQ  are parallel. Since <RBA=RPQ=90-B/2 PQ cuts midpoints of BC, AD, M,N.

Let L be the intersection PQ and BC, ML=1/2 CD, LN=1/2 AB, 

Since M,N are the midpoints of right triangles PBC, QAD

PM=1/2 BC QN=1/2 AD