The perimeter of `Delta ABC` is 36.

In `Delta ABC` we find point D on BC such that the perimeter of `Delta ABD` is the same as the perimeter of `Delta ADC` . Then AB+BD=CD+AC.

We then place point E on AC such that `Delta AEB, Delta BEC` have the same perimeter. Then AB+AE=BC+CE.

Draw the line connecting the point C and the intersection of AD and BE and label the point where this line intersects AB as F.

We are given that BD=10 and CD=2. We are also given that the ratio of BF:AF=3:1. We are asked to determine the perimeter of `Delta ABC` .

Let BF=3x and AF=x.

Then AB=4x, BC=12. AB+10=AC+2 so that AC=AB+8=4x+8.

By this construction AF+AC=BF+BC so that:

x+4x+8=3x+12

5x+8=3x+12

2x=4

x=2.

So AB=4x=8.

BC=12.

AC=4x+8=16.

The perimeter is 8+12+16=36. (See attachment for diagram.)

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