AD and BC are the two bases of trapezoid ABCD, whose side lengths are labelled in the diagram. The extensions of AB and DC intersect at P. Find the perimeter of triangle PAD.
Triangles PBC and PAD are similar....so....
We can find PC as
PC / 15 = (PC + 10) / 27 cross-multiply
27 PC = 15 ( PC + 10)
27PC = 15PC + 150
27PC - 15PC = 150
12PC = 150
PC = 150 /12 = 12.5
And PB = PC
So...perimeter of PAD = 2 (12.5) + 2 (10) + 27 = 72