# Triangles with equal area in a quadrilateral

These are two questions that have come up to my mind. I do not make 2 different forums for the question because the second one is quite short and might be very simple to answer.

1) $$ABCD$$ is cuadrilateral and $$E$$ is the intersection between diagonals $$AC$$ and $$BD$$. Assume that $$\triangle BEC$$ and $$\triangle AED$$ have the same area. Then do $$\triangle ABE$$ and $$\triangle CED$$ have the same area? hat

What I did was to use the formule for the area of a triangle with sides $$a.b$$ and $$c$$ and angle $$A$$ opposite to a:

$$Area = \frac {bc \sin (A)}{2}$$

With that I cant get that: $$BE \cdot EC = AE \cdot ED$$ then with that you can show that $$AB$$ is parallel to $$CD$$ an then triangles $$\triangle ABE$$ and $$\triangle CED$$ have the same area. I posted this question because I thought that there was a simpler way to get to that conclusion.

• One question at a time please, especially since the problems are unrelated. – quasi Oct 13 '18 at 1:30
• 1) is not true, consider trapezoid $ABCD$ – Vasya Oct 13 '18 at 1:33
• isosceles trapezoid – quasi Oct 13 '18 at 1:34
• @quasi: doesn't have to be isosceles – Vasya Oct 13 '18 at 1:39
• @Vasya: True, but if it's isosceles, the counterexample is immediately obvious. In any case, you posted a nice answer. – quasi Oct 13 '18 at 1:45

$$A_{\triangle{ABC}}=A_{\triangle{BCD}}$$ (triangles with the same base between parallel lines).
Hence, $$A_{\triangle{ABE}}=A_{\triangle{CDE}}$$ but $$A_{\triangle{BCE}} \ne A_{\triangle{AED}}$$