A geometry problem by Raman Sharma
ABC is an Equilateral triangle, where AN:NB = 1:2, BL:LC = 1:2 and CM:MA = 1:2
D is the intersecting point of BM and CN E is the intersecting point of AL and BM F is the intersecting point of AL and CN
Find Area(EFD) : Area(ABC)
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1 solution
Let the side of the triangle ABC be 3x
By symmetry,
BM=CN=AL ; BE=CD=AF ; DE=EF=FD triangle DEF is equilateral.
Triangle DMC is similar to triangle CMB as
Angle DCM =MBC and angle DMC is common.
=> DM/x = DC/3x = x/BM
DM = DC/3
Angle MDC =60 [ Angle FDE =60 ]
Applying Cosine law in triangle MDC we get
MC = rt7*DM Area of triangle MDC = 0.5 * 3 * DM^2 * Sin60 = 3rt3/4 * DM^2
Area of triangle ABC = rt3/4 63 DM^2
Area of triangle DEF = 3*3rt3/4 * DM^2
Ratio of DEF/ABC = 1/7