The vertices D, E, F of the rhombus ADEF inscribed in the triangle ABC lie on the sides AB, BC, AC, respectively. Find the segments BE and EC if AB = 14 cm, BC = 12 cm and AC = 10 cm.
Consider all the resulting triangles after placing the ADEF rhombus in the triangle. DE is parallel to AF, and EF is parallel to AD, since ADEF is a rhombus.
Triangles DBE and EFC are similar, so DB / EF = DE / FC = BE / EC. ( 1 ).
Denote AF = X, FC = 10 – x, EF = x, BD = 14 – x,
Substituting the known values. From the similarity of the triangles it follows:
(14 – x) / x = x / (10 – x).
140 – 24 * x + x ^ 2 = x ^ 2.24 * x = 140. x = 35/6.
(14 – x) / x = (14 – 35/6): (35/6) = (14 – 35) / 6): (35/6) = 49/35 = 7/5.
From (1) we take: BE / EC = 7/5.
BE = EC * (7/5).
BE + EC = BC = 12.
EC + EC * (7/5) = (12/5) * EC = 12.EC = 5 (cm), BE = 7 (cm).
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