PROBLEM
In a convex quadrilateral $ABCD$, any triangle determined by a side and the point of intersection of the diagonals has the same area as the adjacent similar triangles. Show that the quadrilateral is a parallelogram.
Drawing
IDEA
Okey, so I'm really confused.
What does adjacent similar triangles? Like, the adjacent similar triangles of $\bigtriangleup AOB$ are $\bigtriangleup BOC$ and $\bigtriangleup AOD$?
If so, I can simply solve the problem, but it seems too easy so I wanna be sure if this is the correct answer.
If so, we can simply see that the area of $\bigtriangleup AOB$ is the height multiplied by $OB$
Then we write that the area of $\bigtriangleup AOD$ is the height multiplied by $OD$, so $OD=OB$.
The same thing with $\bigtriangleup AOB, \bigtriangleup BOC => AO=OC$.
Now we could simply show the requirements.
Hope one of you can help me, telling me if the idea is correct! Thank you!
You are given that each of the four triangles is equal to the two triangles adjacent to it, which are also similar (to each other, not to the first triangle).
Thus for example$$\triangle AOB=\triangle AOD=\triangle BOC$$and$$\triangle AOD\sim\triangle BOC$$But equal triangles under the same height have equal bases. Therefore$$DO=OB$$and$$AO=OC$$And a quadrilateral whose diagonals bisect each other is a parallelogram.
This can further be seen as follows: since $\triangle AOD\sim\triangle BOC$, then$$\angle DAO=\angle BCO$$and hence$$AD\parallel BC$$And since triangles which are equal and similar are also congruent, then$$AD=BC$$This implies (Euclid, Elements I, 33) that $AB$ is equal and parallel to $DC$, and hence that $ABCD$ is a parallelogram.