# Math Help - 3 angle bisectors

1. ## 3 angle bisectors

Code:
               C

B   20    E  15   D    20     F            64             A
Triangle ABC: a = BC, b = AC, c = AB (c > b > a)
CD bisects angle ACB, then CE bisects angle BCD and CF bisects angle ACD.
d = CD, e = CE and f = CF.
a = 40, b = 96, c = 119, d = 30, e = 30, f = 40.
The bisectors divide base AB in integer lengths as shown: 20 + 15 + 20 + 64 = 119.
This is the lowest primitive case of these, lowest being determined by side BC.

What is the next primitive case?
HINT: a <> f and d <> e.

Moderator edit: OK in this subforum.

2. Before I "lose" my solution(!) :
Spoiler:

Triangle sides are 784-2695-2769