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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