IMO-B

Assume Triangle ABC have height- h and base is -b
shaded area will have height is 1/2h ( because D& E are mid point of line AC and BC)
Both shaded area triangle are equal then bas of area is 1/4b
Area of Triangle ABC - 1/2 *bh
Area of both shaded area is -2*1/2*1/4*1/2bh
=1/8bh
Triangle ABC is fraction of shaded area is =(1/2*bh) / (1/8bh)
=1/4 ANS

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Given, Each shaded triangle is similar to △ABC, and the shaded triangles are congruent to one another.

Let's say, h= height of △ABC and b = base of △ABC, further perpendicular at AB from point D is F and point E is H.

Given, △ABC ~ △DFG, So, AC/CB = DF/FG or, h/b = DF/FG

Now, △ABC ~ △ADF. (A-A-A rule)
So, AB/AC = AD/DF or, [Sqrt (h^2 +b^2)]/h = (h/2)/DF

Similarly, △ABC ~ △EHB
So, AB/AC = EB/EH or, [Sqrt (h^2 +b^2)]/h = (b/2)/EH, So, (h/2)/DF= (b/2)/EH.

since △DFG congruent to △EGH. So, DF= EH
now, h = b and DF = h/(2*sqrt2)

So, (area of △DFG + area of △EGH) / area of △ABC = [1/2*(DF*FG)* 2] / [1/2*(AC*BC)] = 1/4.

I think B.

Answer is b
As we know all will form right angle triangle
And whole triangle is divided in 8 parts out of which 2 is shaded
Draw a line perpendicular from c

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