Elegant collinearity is in my intuition(Self-made)

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Fm Jakaria
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Elegant collinearity is in my intuition(Self-made)

Unread post by Fm Jakaria » Thu Aug 28, 2014 11:14 pm

Let 123 be a triangle, in which medians from 1,2,3 respectively intersect it's circumcircle again at 4,5,6. The tangents to this circumcircle at 4,5,6 respectively intersects 23,31,12 at 7,8,9. Prove that 7,8,9 are collinear.
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nishat protyasha
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Re: Elegant collinearity is in my intuition(Self-made)

Unread post by nishat protyasha » Fri Aug 29, 2014 5:58 pm

I have placed $$A,B,C,D,E,F,G,H,I$$ instead of $$1,2,3,4,5,6,7,8,9$$ for my convenience. Now the question is :
Let $$ABC$$ be a triangle, in which medians from $$A,B,C$$ respectively intersect its circumcircle at $$D,E,F$$. The tangents to this circumcircle at $$D,E,F$$ respectively intersects $$BC,AC,AB$$ at $$G,H,I$$. Prove that $$G,H,I$$ are collinear.
Proof :
$$\triangle{BDG} \sim \triangle{DCG}$$.
So, $$\dfrac{BD}{DC}= \dfrac{BG}{DG}= \dfrac{DG}{CG}$$
Let $$M$$ be the midpoint of $$BC$$.
So $$(ABM) = (AMC)$$
$$\Rightarrow \dfrac{1}{2} * AB*AM * sin\angle{BAD} = \dfrac{1}{2} * AM*AC *sin\angle{DAC}$$
$$\Rightarrow \dfrac {sin\angle BAD }{sin\angle DAC }= \dfrac{AC}{AB}$$
In $$\triangle{BDC}$$,
$$ \dfrac {BD}{sin\angle BCD }= \dfrac {DC}{sin\angle DBC }$$
$$\Rightarrow \dfrac {sin\angle BCD }{sin\angle DBC }= \dfrac{BD}{DC}$$
Quad $$ABCD$$ is concyclic.
So, $$\angle{BAD} = \angle BCD $$ and $$\angle{DAC} = \angle DBC $$
So, $$\dfrac{AC}{AB} = \dfrac {sin\angle BAD }{sin\angle DAC }= \dfrac {sin\angle BCD }{sin\angle DBC }= \dfrac{BD}{DC} = \dfrac{BG}{DG}= \dfrac{DG}{CG}$$
$$\dfrac{BG}{DG} = \dfrac{AC}{AB}$$
$$\dfrac{DG}{CG} = \dfrac{AC}{AB}$$
So, $$\dfrac{BG}{DG} * \dfrac{DG}{CG} = \dfrac{AC^2}{AB^2}$$
$$\Rightarrow \dfrac{BG}{CG} = \dfrac{AC^2}{AB^2}$$
Like this,
$$\dfrac{AI}{BI} = \dfrac{BC^2}{AC^2}$$ and $$\dfrac{CH}{AH} = \dfrac{AB^2}{BC^2}$$
Now, $$G,H,I$$ are points on $$BC,AC,AB$$ respectively of $$\triangle{ABC}$$.
$$\dfrac{AI}{BI} * \dfrac{BG}{CG} * \dfrac{CH}{AH} = \dfrac{BC^2}{AC^2} * \dfrac{AC^2}{AB^2} * \dfrac{AB^2}{BC^2} = 1$$.
So, $$G,H,I$$ are collinear.

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