<Tri-2025-2278>, hint

Let H be the orthocenter , A' feet of altitude of AA', B' feet of altitude of BB'.

Let C" intersection of circles of diameters  HA', HB, respectively.

Since <HC"A'=<HC"B'=90, A', C", B' are collinear

By angle chasing <C'A'B=<A=<BHA', thus <HA'B"=<HC"B"=90-A.

Similarly <HC"A"=90-B. Hence <B"C"A"=90-A+90-B=180-(A+B)=C

Similarly <C"A"B=A, <A"B"C"=B

Hence triangles ABC and A"B"C" are similar.