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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.

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Did you mean FECD? – Steven Gregory Jul 27 '15 at 2:01
@StevenGregory, edited, thanks. – rah4927 Jul 27 '15 at 13:12

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 $$\angle DCE=\angle DCA=\tan^{-1}\left(\frac{AD}{CD}\right)=\tan^{-1}\left(\frac{3}{2}\right)\approx 56.31^{\circ}\\ \angle EBC=\angle FBC=\tan^{-1}\left(\frac{DF}{BD}\right)=\tan^{-1}\left(\frac{1.5}{2}\right)\approx 36.87^{\circ}$$

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Hmm,I should have mentioned not to use trig as I am just learning them.But thanks for the answer – rah4927 Jul 27 '13 at 12:05
Also,what do we use to find the relevant angles? – rah4927 Jul 27 '13 at 12:08
Ok. You'll need trigonometry to find out the angles, I'll put the formulas to find out them, you can later on check it out yourself. – Samrat Mukhopadhyay Jul 27 '13 at 12:14
You mean angle DCE equals angle DBA or ACD. – rah4927 Jul 27 '13 at 12:31
Yes @rahul, that's what I mean. – Samrat Mukhopadhyay Jul 27 '13 at 12:33

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Let $\mathbf{ABC}$ represent the area of triangle $ABC$

$D = \frac 12B + \frac 12C$

$F = \frac 12(A + D) = \frac 12A + \frac 14B + \frac 14C$

$E=(1-\alpha) A + \alpha C$
$E = (1-\beta)B + \beta F = \frac 12 \beta A + (1 - \frac 34\beta)B + \frac 14 \beta C$

$(1-\alpha) A + \alpha C = \frac 12 \beta A + (1 - \frac 34\beta)B + \frac 14 \beta C$
$\alpha = \frac 13$
$\beta = \frac 43$

Let $\mathbf X, \mathbf Y$, and $\mathbf Z$ be the areas of the indicated regions.

The two triangles with areas marked as $\mathbf Z$ have equal areas because their bases are equal and they have the same altitude. Clearly

$$\mathbf Z = \frac 14 \mathbf{ABC}$$

Since $\mathbf E = -\frac 13 \mathbf B + \frac 43 \mathbf F$.
then $\mathbf F = \frac 14 \mathbf B + \frac 34 \mathbf E$.
This implies that $BF = \frac 34 BE$, or
$BF = 3FE$. It follows that
$$\mathbf Y = \frac 13 \mathbf Z = \frac{1}{12} \mathbf{ABC}$$

$$\mathbf X = \frac{5}{12} \mathbf{ABC}$$

The area of $\mathbf{FECD} = \dfrac{5}{12}$ the area of triangle $ABC = \frac 52$.

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