On the side KM of triangle KMN, point L is chosen so that KL: LM = 2: 5. Through point L, parallel to the sides KN

On the side KM of triangle KMN, point L is chosen so that KL: LM = 2: 5. Through point L, parallel to the sides KN and MN, segments are drawn, the ends of which lie on the sides MN and KN. Find the lengths of these segments if MN = 7, KN = 21.

Let a triangle KMN be given, on the side KM of which the point L is chosen so that KL: LM = 2: 5. We introduce the proportionality coefficient k, then KL = 2 ∙ k; LM = 5 ∙ k; KM = KL + LM; КМ = 2 ∙ k + 2 ∙ k = 7 ∙ k. The sides of the triangle are MN = 7, KN = 21. Moreover, through the point L parallel to the sides KN and MN segments are drawn, the ends of which lie on the sides MN and KN – points B and A, respectively. We get that Δ LMВ ∽ Δ KМN and Δ KL А ∽ Δ KМN since they are formed by straight lines parallel to the sides of the triangle. The corresponding sides of such triangles are proportional:
LM: KM = LB: KN and
KL: KM = KA: KN;
or (5 ∙ k) / (7 ∙ k) = LВ / 21 and
(2 ∙ k) / (7 ∙ k) = LA / 7.
Having solved the proportions, we get:
LB = 15;
LA = 2.
Answer: the lengths of the formed segments are LВ = 15; LA = 2.