How can I prove this?

Question

Let P be any point in the interior of a rectangle ABCD. Four triangles are formed by joining P to each vertex A, B, C, and D.

How can I prove that:
Area of Triangle APD + Area of Triangle BPC = Area of Trianble APB + Area of Triangle PCD?

Answer 1

Draw the line perpendicular to AD through P and let it intersect the rectangle at E and F, with E on AD and F on BC. Then APD has base AD and height EP, and BPC has base BC and height FP. Therefore, the total area of APD and BPC is AD*EP/2 + BC*FP/2. However, since AD and BC are opposite sides of a rectangle, AD=BC, so we have that the total area is 1/2 AD (EP+FP). Now, since P is between E and F, EP+FP=EF, which is congruent to AB, so this is 1/2 AD*EF. However, DEFC is a rectangle, so EF = DC, so the total area of those two triangles is 1/2 AD*CD, which is 1/2 Area(ABCD). By constructing the line perpendicular to CD through P and applying this same reasoning, you can conclude that Area(APB) + Area(CPD) = 1/2 Area(ABCD) as well, so that the equality Area (APD) + Area (BPC) = Area (APB) + Area (CPD) then follows from transitivity.