In trapezoid ABCD with bases AD = 8 cm and BC = 3 cm, point K is the middle of AD. Diagonal AC meets line segment BK at point M. Find BM: MK.
Since point K is the middle of the base AD, then AK = DK = AD / 2 = 8/2 = 4 cm.
The BMC and AMC triangles are similar in two angles. Angle ВМС = АМК as vertical angles at the intersection of straight lines АС and ВМ, angle СМ = ВКА as criss-crossing angles at the intersection of parallel straight lines ВС and AK secant ВК.
Let’s determine the coefficient of similarity of triangles. K = BC / AK = 3/4.
Then BM / MK = 3/4.
Answer: Parties BM and MK are treated as 3/4.
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