Let's pretend that we know the values of all the sides of the small triangle \((a,b,c)\). We have that statement that \[r=\frac ad = \frac be = \frac cf\]We'll explore just the first part of that expression.
\[r = \frac ad\]Multiply both sides by \(d\) and we get
\[dr = a\]Divide by \(r\) and we have
\[d = \frac ar\]Divide by \(a\) and we have
\[\frac da = \frac 1r\]
In other words, the ratio of side \(d\) to side \(a\) is \(1/r\).
Now let's do that for the other sides, and we get
\[e = \frac br\]\[f = \frac cr\]
Do you see the pattern? Each of the sides of the sides of the big triangle can be gotten by multiplying the corresponding side of the small triangle by a constant, \(1/r\).
Now the perimeter of the small triangle is just the sum of its sides, \((a+b+c)\). What is the perimeter of the big triangle? Again, it is the sum of its sides, \((d + e + f)\).
What happens if you write the perimeter of the big triangle using those expressions we got for \(d, e, f\)?
\[(\frac ar + \frac br + \frac cr)\]Looks pretty interesting! What if we factor out \(1/r\)? (This is just using the distributive property of multiplication in reverse)
\[\frac1r(a+b+c)\]but \((a+b+c)\) is just the perimeter of the small triangle, and that factor of \(1/r\) in front is just the ratio of any side in the small triangle to the corresponding side in the big triangle!
So, we've established that if the sides are proportional, the perimeter is proportional as well.