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Question

Let ABC be an acute angle triangle in which D, E, F are points on BC, CA, AB respectively such that AD BC, AE = EC and CF bisects 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 3 = 1.73]
733026_5ab47bb68e6d4cbfa7328357cb3dd677.png

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Solution

Given, FN=NC=3

In ACF , CEAE=CNNF=1.

So, by midpoint theorem, EN||AFDE||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, DNAF=MNMF=12.

DN=EN=AF2

In CED, by angle bisector theorem

CECD=ENDN

But EN=DN

CD=CEAB=AC

ABC is an equilateral triangle with altitude FC=6

AB=BC=AC=CF×23=123=43

Perimeter of ABC=3×43=123=20.76

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