A line parallel to side AC of triangle ABC meets sides AB and BC at points M and N, respectively.

A line parallel to side AC of triangle ABC meets sides AB and BC at points M and N, respectively. Find BN if MN = 20, AC = 35, NC = 39.

Since MN is parallel to AC, it cuts off a similar triangle MBN from the original triangle. On the third basis of similarity:
BC / BN = AC / MN.
The BC side consists of two segments BN and NC:
BC = BN + NC.
Then:
(BN + NC) / BN = AC / MN.
Substitute the known values and find the length BN:
(BN + 39) / BN = 35/20;
35BN = 20 (BN + 39) (proportional);
35BN = 20BN + 780;
35BN – 20BN = 780;
15BN = 780;
BN = 780/15;
BN = 52 (conventional units).
Check: (BN + NC) / BN = AC / MN;
(52 + 39) / 39 = 35/20;
91/39 = 35/20;
1.75 = 1.75.
Answer: BN = 52 conventional units.