Question

Let ABC be an acute angle triangle in which D, E, F are points on BC, CA, AB respectively such that AD $\perp$ BC, AE = EC and CF bisects $\angle$C internally. Suppose CF meets AD and DE in M and N respectively. if FM = 2, MN = 1 and NC = 3. Then the perimeter of $△$ ABC is [use $\sqrt{3}$ = 1.73]

Answer 1

Given, $FN=NC=3$

In $△ACF$ , $\frac{CE}{AE}=\frac{CN}{NF}=1$.

So, by midpoint theorem, $EN||AF⟹DE||AB$.

In $△ABC,CE=EA$ and $DE||AB$.

So, by intercept theorem $BD=DC$

$⟹ABC$ is an isosceles triangle with $AB=AC$

Consider triangles $DMN$ and $AMF$.

They are similar to each other by $AA$ similarity criterion.

So, $\frac{DN}{AF}=\frac{MN}{MF}=\frac{1}{2}$.

$\therefore DN=EN=\frac{AF}{2}$

In $△CED,$ by angle bisector theorem

$\frac{CE}{CD}=\frac{EN}{DN}$

But $EN=DN$

$\therefore CD=CE⟹AB=AC$

$△ABC$ is an equilateral triangle with altitude $FC=6$

$\therefore AB=BC=AC=CF\times \frac{2}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}$

Perimeter of $△ABC=3\times 4\sqrt{3}=12\sqrt{3}=20.76$