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

REF.Image.

Given :- $AD\perp BC$

$AE=EC$

$FM=2$

$MN=1$

$NC=3$

$E,N$ or mid point of AC, CN

$\Delta ADC;$ $EN\parallel AF$

$\angle EDC=\angle B.$

$△ADC;$

$EC=EA=DE$

$\angle EDC=\angle C$

$\angle C=\angle B$

then $AB=AC.$

AD Bisect base BC.

Since AD divide CF such that

$CM:MF=2:1$

CF is median , point -F is mid point of AB.

$EF\parallel BC$

$EF=\frac{BC}{2}$

$\angle EFC=\angle FCB=\angle ECF$

$EF=EC=AC/2$

$BC=AC=AB.$

$△ABC$ Becomes equilateral triangle

median = Angle Bisector = Altitude

$CE=6$

$CE=\frac{\sqrt{3}}{2}\times AB$

$AB=\frac{12}{\sqrt{3}}$

Perimeter $=12\sqrt{3}$ unit