Triangle Mystery
In the triangle ABC, then base BC is trisected at D and E. The line through D, parallel to AB meets AC at F and line through E parallel to AC meets AB at G. Let EG and DF intersect at H. What is the ratio of the sum of the area of parallelogram AGHF and the area of the triangle DHE to the area of the triangle ABC?
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1 solution
triangle FDC is similar to triangle ABC. DC = (2/3) BC, so area [FDC] = (4/9) [ABC] triangle BGE is similar to triangle ABC. BE = (2/3) BC, so area [BGE] = (4/9) [ABC] triangle DHE is similar to triangle ABC. DE = (1/3) BC, so area(DHE] = (1/9) [ABC] Parallelogram AGHF = [ABC] - { [BGE] + [DFC] -[DHE]} = [ABC] - {4/9 + 4/9 - 1/9} [ABC] = (2/9) [ABC] {parallelogram AGHF} + [HED] = (2/9) [ABC] + (1/9) [ABC] = (1/3)*[ABC]