Jatheen
Last Activity: 6 Years ago
Given FN = NC = 3. In triangle ACF , CE/AE = CN /NF = 1.
So, by midpoint theorem, EN || AF. => DE || AB. and EN = AF/2
In triangle 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, DN/AF = MN/MF = ½. Therefore DN = EN = AF/2.
In triangle CED, by angle bisector theorem CE/CD = EN/DN. But EN = DN.
Therefore CE = CD => AC= BC. So, ABC is an equilateral triangle with altitude CF = 6.
So, perimeter = 3 * AB = 3 * 2* CF/ sqrt(3). = 6* 6/ sqrt (3) = 12* sqrt(3)= 20.76