Question

Let ABC be an acute angled 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 AB and DE in M and N respectively. If FM$=2$, MN$=1$, NC$=3$, Find the perimeter of the triangle ABC.

Answer 1

$\overline{FN}=\overline{NC}=3$ and $\overline{AE}=\overline{EC}$ (given)

So, According to SAS Similarity,

$△NEC\sim △FAC$

So, $\overline{NE}||\overline{AF}\Rightarrow \overline{AB}||\overline{ED}$

ie, $\overline{AD}$ is a perpendicular bisector

$\Rightarrow △ABC$ is an isosceles triangle.

Also, $\overline{FC}$ passes through centroid because $\overline{FM}:\overline{MC}=2:1$

So, M is the centroid and incentre of the triangle.

ie, $\overline{FM}=\overline{MD}=2=r$ (inradius)

And $\overline{AD}=6$

$\overline{DC}=\sqrt{{\overline{MC}}^2-{\overline{MD}}^2}=\sqrt{4^2-2^2}=2\sqrt{3}=\frac{1}{2}\overline{BC}$

$Area=\frac{1}{2}\overline{AD}\times \overline{BC}=6\times 2\sqrt{3}=\frac{1}{2}r\left(Perimeter\right)$

So, $Perimeter=12\sqrt{3}$