Elegant collinearity is in my intuition(Self-made)
- Fm Jakaria
- Posts:79
- Joined:Thu Feb 28, 2013 11:49 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.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.
- nishat protyasha
- Posts:33
- Joined:Tue Sep 17, 2013 12:02 am
- Location:Sylhet, Bangladesh.
Re: Elegant collinearity is in my intuition(Self-made)
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.
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.