It seems to wrong statement but how can we prove it? Any hint will be appreciated.
Area of triangles inside quadrilateral.
areaeuclidean-geometrygeometryquadrilateral
Related Solutions
The area of the quadrilateral will be $$\Delta BCE-\Delta BDF$$ Now, you know $BD=2,\angle BDF=90^{\circ}$ and you can calculate $AD$ to be $3$ so that $DF=1.5$. Then you can calculate the area of $\Delta BDF$. Now, you can calculate $\angle DCE$ and $\angle EBC$ to apply the $\sin$ theorem to get $CE$. Then you can get the area of $\Delta BCE$ by $\displaystyle \frac{1}{2}BC. CE \sin \angle BCE$, and you are done.
Relevant Calculations $$\angle DCE=\angle DCA=\tan^{-1}\left(\frac{AD}{CD}\right)=\tan^{-1}\left(\frac{3}{2}\right)\approx 56.31^{\circ}\\ \angle EBC=\angle FBC=\tan^{-1}\left(\frac{DF}{BD}\right)=\tan^{-1}\left(\frac{1.5}{2}\right)\approx 36.87^{\circ}$$
Oops. I forgot to put the distance in the picture. post to brook is $500$, $m\angle Post-brook-mill = 135$ (going from NE to due E). Brook to old mill is $300$. $m\angle brook-mill-wiggins=105$ (going from due east to south ($90$ and eat by $15$ degrees). And old mill to wiggins rd is $200$.
So .... have at it.
Best Answer
We can-not prove it because it's wrong in the general.
But we can get any quadrilaterals with this property.
Indeed, let $S_{\Delta AOB}=a,$ $S_{\Delta COB}=b,$ $S_{\Delta COD}=c$ and $S_{\Delta AOd}=d,$
Thus, $$\frac{d}{a}=\frac{DO}{BO}=\frac{c}{b},$$ which gives $$ac=bd.$$ Now, let $a>d$.
Thus, $b>c$ and $$(a+c)^2=(b+d)^2$$ or $$a^2+2ac+c^2=b^2+2bd+d^2$$ or $$a^2-2ac+c^2=b^2-2bd+d^2$$ or $$(a-c)^2=(b-d)^2$$ or $$(a-c+b-d)(a-c-b+d)=0$$ or $$a-c=b-d$$ and since $$a+c=b+d,$$ we obtain: $$a=b$$ and $$c=d.$$ Id est, in our quadrilateral $O$ is a mid-point of $AC$
or by the same way we can show that $O$ is a mid-point of $BD$ (if $a>b$, for example).