the algebra solution is correct. the diagram solution is not

correct, because the triangles with the same color are not congruent, only if point p is the center of the square.

Let ABCD be a square, and P an arbitrary point inside the square. Prove that the sum of the areas of triangles ABP and CDP is the same as the sum of the areas of triangles BCP and DAP. There is a solution "without words" that uses a single diagram. C Area BAP + Area CDP = BA: 2 + CD.h, B We do the same with the sides AD and BC, for triangles BCP and DAP Area BCP + area Ac. Hy + AD .Hy As ABCD is a square , we have : BA-CD = BC = AD = I, So replacing we have : Area BAPt Area CDP = "2 + Like = 1. (hithe) Area Bop+ Areu DAP = .he + 1. hy= 1( to thy) And finally h, the = hy + hy = 1 so we obtain Area BAP + Area CDP = 1.1 = 1 Area BCP+ Area DAP = 1.1 = 1 A So we have: Area BAP+ Area CDP = Area BCP + Area DAP Yes, the triangle is the same as the triangle area, since each of them has a traingle of the same area. You'll note that the four different colors that are formed are the same in the diagram