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 AD and DE in M and N respectively. If FM = 2, MN = 1, NC = 3, Find the perimeter of the triangle ABC.

Answer 1

Here $\overline{FN}=\overline{NC}=3$ and $\overline{AE}=\overline{EC}$ (Given),

ie,$\overline{NE}\parallel \overline{AF}$

$\Rightarrow \overline{AB}\parallel \overline{ED}$ (Extended Line segment)

Then, $\angle DEC\cong \angle BAC$

So, $\overline{BD}=\overline{DC}$ ($\because △ABC\sim △EDC$)

i.e. Perpendicular bisector $\overline{AD}$ is also the median of $△ABC$

So, $△ABC$ is an isosceles triangle with base $\overline{BC}$.

Then, $\overline{FC}$ is also a median of the triangle and $\overline{AD}$ is also the angle bisector of $\angle A$.

ie, H is both the centroid and incentre of the triangle.

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

and, $\overline{MA}=2\overline{MD}=4\Rightarrow \overline{AD}=6$ (Centroid divides the median in $2:1$ ratio)

In $△DMC$,

${\overline{MD}}^2+{\overline{DC}}^2={\overline{CM}}^2\Rightarrow \overline{DC}=\sqrt{16-4}=2\sqrt{3}$ (Pythagoras Theorem)

$\Rightarrow \overline{BC}=4\sqrt{3}$ (Divided by median)

and $r=\frac{△}{s}\Rightarrow s=\frac{△}{r}\Rightarrow \frac{1}{2}(a+b+c)=\frac{1}{2}\frac{\overline{AD}\overline{BC}}{r}$

So, Perimeter $=(a+b+c)=\frac{6\times 4\sqrt{3}}{2}=12\sqrt{3}$