A line parallel to side AC of triangle ABC intersects sides AB and BC at points K and M, respectively. Find AC if BK: KA = 1: 4, KM = 13.

bystudin | May 31, 2020 | Education

Consider the triangles ABC and KBM.
∠B is common.
∠BAC = ∠BKM (since these are the corresponding angles)
∠BCA = ∠BMK (since these are also the corresponding angles)
Therefore, these triangles are similar in the first sign of similarity.
Then by the definition of such triangles:
BA / BK = AC / KM
(BK + KA) / BK = AC / KM
1 + KA / BK = AC / KM
1 + 4/1 = AC / KM
5 = AC / 13
AC = 5 * 13 = 65
Answer: AC = 65

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