<Tri-2025-2203>, hint

 Suppose BM and BN cut AC in M' and N' respectively

Since triangles BDQ and BEP are isosceles <BND=<BME=90

BN=AC cosA tan(A/2) cos(A/2), BM=AC sinAtan(C/2) cos(C/2)

BN'=BH(altitude)/ cos(A/2)   BM'=BH/ cos(C/2)

BM/BN=BM'/BN'  <----(cosA sin A/2)/(cosC sinC/2)=(cosC/2)/(cosA/2)

cosAsinA=cosCsinC