In triangle ABC the heights AD and BF are drawn, which meet at point G. Prove that triangle AGF is similar to triangle BCF.

First, we prove that the triangle BFC is similar to BDG1. Angle GDB = BFC = 90 ° since BF and AD are heights 2. Angle B is common for these triangles. We conclude that triangles are similar. Next, we prove that triangle AGF is similar to BGD1. Angle GDB = BFA = 90 ° as BF and AD are heights 2. Angle AGF = DGB since these are vertical angles and they are equal We conclude that triangles are similar If BFC is similar to BDG and AGF is similar to BGD, then AGF is similar to triangle BCF.