Prove Parallelogram ABCD: Triangle APD = ABP + DCP

[jtf2eh]

Parallelogram proof :(

Homework Statement

In parallelogram ABCD, P is any point on BC. Prove that triangle APD = triangle ABP + triangle DCP

Homework Equations

n/a

The Attempt at a Solution

I really don't know where to start :(

any help would be greatly appreciated :)

 

[HallsofIvy]

First do you mean that the areas satisfy that equation? If so just note that the lengths of the bases of the two triangles ABP and DCP add to the length of the base of triangle APD and all three triangles have the same height.