Let us consider an isoscles triangle ABC where $AB=AC=\sqrt{13}$,and length of $BC=4$.The altitude on $BC$ from $A$ meets $BC$ at $D$.Let $F$ be the midpoint of $AD$.We extend $BF$ such that it meets $AC$ at $E$.We need to calculate the area of quadrilateral $FECD$.
Just so you know,I made this problem up myself.I have tried to calculate areas of other parts of the triangle but I cannot find the area of triangle $AFE$ or $BAE$.We know the area of triangle $ABD$=$3$,and therefore of $AFB$ and $BFD$ each is $1.5$.But I cannot see where we can go from here.I am also interested in solutions involving no trigonometry.
Any help will be appreciated.
Best Answer
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 $$\angle DCE=\angle DCA=\tan^{-1}\left(\frac{AD}{CD}\right)=\tan^{-1}\left(\frac{3}{2}\right)\approx 56.31^{\circ}\\ \angle EBC=\angle FBC=\tan^{-1}\left(\frac{DF}{BD}\right)=\tan^{-1}\left(\frac{1.5}{2}\right)\approx 36.87^{\circ}$$