Triangle ABC is right-angled at A, and AB < AC. A point H is on the segment AC (H is different from A and C). Denote E as the perpendicular projection of H onto BC. Prove that angle HBC = angle EAC.

Yes, but we are supposed to solve it without any knowledge about cyclic quads or anything about circles at all. We were supposed to solve it using knowledge about the proportionality theorem on triangles and the three cases of similar triangles. Can you try to do another solution for that?

Quad ABEH is cyclic (opposite <'s add to 180 deg)

Therefore <HBE = <EAH (Inscribed Angle Theorem)

Since both BEC and AHC are straight, then <HBC = <EAC

Yes, but we are supposed to solve it without any knowledge about cyclic quads or anything about circles at all. We were supposed to solve it using knowledge about the proportionality theorem on triangles and the three cases of similar triangles. Can you try to do another solution for that?

What exactly is the "Proportionality theorem" ?