can be done by extending the side AC (b) to D such that ad=ab and by similarity in triangles ABC and BDC,we get AC/BC=BC/DC.thus a/b=b+c/a and a^2=b(b+c)
I have seen it somewhere before.Maybe an old rmo question?
Well as you have very little time left, I would suggest visiting mathlinks.ro
This is an international forum is most used by students preparing for math olympiads across the world
There is a large bank of questions from contests across the world over several years which you will find very useful
The angles A,B,C of the triangle are
So the questions wants us to prove that
But this is true as
one question for your preparation,In a triangle ABC, angle A is twice angle B. Show thata^2 = b · (b + c).
Can i try it? -> thanks
The question can be represented in the figure above....!!
We join vertex A with a point D on BC such that AD becomes the angle bisector of BAC
Now, by angle bisector theorem
c/b = m/n
=> c/b + 1 = m/n + 1
=> (b + c)/b = (m + n)/n
=> (b + c)/b = a/n ------- (i)
One more thing can be noted is
triangle ABD is isocelous
=> AD = BD = m
Again, LADC = x + x = 2x
or, LADC = 2xSo we see triangle ABC and DAC are similar (By AAA - criterion)
=> AB/AD = BC/AC = AC/CD=> c/m = a/b = b/n
or, a/b = b/n
=> n = b2/a
putting, the value of n in (i) we get,
(b + c)/b = a/b2/a
=> a2 = b.(b + c) proved..!!
Well if it was a question from RMO (then this is the first time that i could solve one on my own ha ha...!!)
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