Question

ff

Answer 1

It is usually easier to solve these problems when they are stated correctly. First of all, P is irrelevant. More importantly, what we are trying to show is that HE=EF=FG.

In a triangle, if a line parallel to the base intersects the triangle, the length of the intersecting segment is the base times (h/H) where H is the altitude of the triangle and h is the distance of the line to the triangle's base. Hence, since ADB and ACB are triangles of the same altitude with a common base, HE=FG.

By the same token, since ADM and ACM are triangles with the same altitude and equal bases (on the same line), any line parallel to their base cuts them equally. Hence HE=EF. QED

Answer 2

HUH?

Answer 3

so AB

Answer 4

yes