Inradius = [ ABC ] / [semiperimeter of ABC]
ABC is a right triagle so the area = (1/2)(product of the legs) = (1/2) (5 * 12) = 30
Semi-perimeter = [ 5 + 12 + 13 ] / 2 = 15
Inradius = (30) / ( 15) = 2
Because ID is parallel to AC, then AE = AC - ID = 12 - 2 = 10
Draw AI.....this bisects angle A
And IE = 2
And triangle AIE is right because angle AEI = 90°
So....the area of AEI = (1/2)(2)(10) = 10
And by SAS, triangle AFI is congruent to triangle AEI....so its area also = 10
So [ AEIF ] = 10+ 10 = 20
