Answer:
50 cm
Step-by-step explanation:
Consider isosceles triangle AEF in the attachment. Point B is the center of the incircle, and it divides altitude AC into segments having the ratio 12:5.
Triangle ABD is similar to triangle AEC by AA similarity. (Angle A is the same for both right triangles. Then the ratio of hypotenuse to short leg will be the same for each. In triangle ABD, that ratio is 12:5, as given by the problem statement. Since we know AE = 60 cm, also from the problem statement, we know that ...
AB/BD = AE/EC
12/5 = 60 cm/EC
so ...
EC = (60 cm)·(5/12) = 25 cm
Base length EF is twice that, or 50 cm.
