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.

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

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Claim 1) is not true. Consider trapezoid.
$A_{\triangle{ABC}}=A_{\triangle{BCD}}$ (triangles with the same base between parallel lines).
Hence, $A_{\triangle{ABE}}=A_{\triangle{CDE}}$ but $A_{\triangle{BCE}} \ne A_{\triangle{AED}}$


2) it is true because

every edge of the regular polygon is a chord of its circumcircle and a tangent to its incircle with the point of tangency being its midpoint.


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