In $\Delta ABC$ has area $120$ with $AB = 50, AC= 10.$ Let $D,E$ be the midpoints of $AB,AC$ respectively.

Question

In $\Delta ABC$ has area $120$ with $AB = 50, AC= 10.$ Let $D,E$ be the midpoints of $AB,AC$ respectively. The angle bisector of $\angle BAC$ intersects $DE$ and $BC$ at $F,G$ respectively. Find the area of quadrilateral $FDBG$.

What I Tried: Here is a picture :-

All the results which I got are in the picture with the appropriate use of Midpoint Theorem as well as Angle Bisector Theorem. Since we are given the area, I can use it to find $BC$ and hence find $x$ . But how am I going to find the area of the trapezium $BDFG$ ? Also, finding $BC$ using Heron's formula will be a bit difficult algebra, I am wanting for some easy methods here.

Can anyone help? Thank You.

Answer 1

Say $\triangle AFE = a$

Then $\triangle ADF = 5a$ (same height as $\triangle AFE$ but base is $5$ times).

$\triangle AGC = 4a$ (both base and height are double of $\triangle AFE$).

$\triangle ABG = 5 \times \triangle AGC = 20a$ ($\triangle AFE$ and $\triangle AGC$ share the same height but base ratio is $5:1$).

Adding them up $\triangle AGC + \triangle ABG = 4a + 20a = 120 \implies a = 5$.

So area of quadrilateral $= \triangle ABG - \triangle ADF = 20a - 5a = 75$.

Answer 2

$$[FDBG] = (1-\tfrac{1}{4})[ABG]= \dfrac{3}{4} \cdot \dfrac{5}{6} [ABC] $$

Since $[ADF] = \tfrac{1}{4}[ABG]$. And $\triangle ABG$ has same height but base $5/6$ th of $\triangle ABC$

So $$[FDBG] = 75$$

Answer 3

We have $$ \begin{aligned}{} [FDBC] &=[ABG]-[ADF]=[ABG]-\frac 14[ABG]=\frac 34[ABG]\\ &=\frac 34\cdot\frac {BG}{BC}[ABC]= \frac 34\cdot\frac {AB}{AB+AC}[ABC] \\ &=\frac 34\cdot\frac{50}{50+10}\cdot 120 \ . \end{aligned} $$