A straight line parallel to the side AC of the triangle ABC intersects the sides AB and BC at points M and N

A straight line parallel to the side AC of the triangle ABC intersects the sides AB and BC at points M and N, respectively. Find BN if MN = 16, AC = 20, NC = 15.

Decision.
The straight line MN, parallel to the side AC of the triangle ABC and intersecting its sides, cuts off a similar triangle MBN from it according to the first attribute, since the angles formed by the straight lines MN || AC and secant sides AB and BC will be equal as respectively. Line MN intersects sides AB and BC at points M and N, respectively, so AC: MN = BC: BN. Let the length of the segment BN = x, then the length of the segment BC = x + 15, since it is known from the condition of the problem that NC = 15. Knowing that MN = 16, AC = 20, we compose the equation:
20: 16 = (x + 15): x;
20 ∙ x = 16 ∙ (x + 15);
X = 60.
Answer: the length of the segment is BN = 60.