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Any help will be appreciated.

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Did you mean FECD? –  Steven Gregory Jul 27 at 2:01
@StevenGregory, edited, thanks. –  rah4927 2 days ago

2 Answers 2



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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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