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vcbabu
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IMO Area od ABC : Area of FBC = 4:1

My approach was....

Let Area of Abc be X.

=> Area of BDC = x/2 (since BD divides ABC into 2 equal halves)

=> Area of BFC = x/4 (since BF divides BDC into 2 equal halves)

Therefore Area of ABC : Area of BFC = x:x/4 = 4:1

Wats the OA??
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Pls. post the OA.This is a good and a tricky question.
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The answer is 4:1 only if AB = BC. If you divide the triangle in half ABD and DBC share one side, DB, while one side of each triangle, AD & DC, has the same lenght. The ratio of the remaining side of each triangle, AB and BC, can have an infinite number of values. Since you can compute the area of a triangle using the lenght of all three sides the ratio of the area the two triangles must have an infinite number of values, depending of what kind of triangle that is originally used in the question (which is not mentioned here).

Please take note that English is not my native language :)
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Hi,

I will try to explain my solution:

Since DE connects the centers of AC and BC then it is parralel to AB and is 1/2 AB. Then Triangle DEC has area 1/4 of the area of ABC.
Now Triangle DFC has the same area as triangle FCE since DF=FE and the height is same.
Triangle FCE has the same area as triangle FBE since BE=EC and height is the same.
So triangles DFC, FCE and FBE have same areas.
Then the area of FBC= area DCE and the ratio is 4/1

Regards



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