anirudh215
Oct17-09, 03:26 AM
I posted this in the geometry subforum, but I think that might have been the wrong place to post this.
Given two triangles with vertices A1, B1, C1 and A2, B2, C2 respectively. A1A2, B1B2, C1C2 are extended to meet at a point V say. Now, B1C1 and B2C2 are extended to meet at L, A1B1 and A2B2 meet at N and A1C1 and A2C2 meet at M. Prove that L, M and N are concurrent.
Proof (as given in text):
Let A1B1C1 be the reference triangle and V be the unit point (1,1,1). A2 is on the join of A1(1,0,0) and V(1,1,1), so it can be taken as (1+p,1,1). Similarly, the point B2 is given by (1,1+q,1) and C2 by (1,1,1+r).
Now, the line B2C2 is
\left|\stackrel{\stackrel{x}{1}}{1}\stackrel{\stac krel{y}{1+q}}{1}\stackrel{\stackrel{z}{1}}{1+r} \right| = 0.
The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0
i.e. x=0, \frac{y}{q} + \frac{z}{r} = 0
Therefore, L lies on the line \frac{x}{p}+ \frac{y}{q}+ \frac{z}{r} = 0. By symmetry, so do M and N.
Hence proved
From start to finish, I can't get it. Can someone please explain to me what all this means?
For the diagram that goes with the proof, please refer to my topic in the geometry section
http://www.physicsforums.com/showthread.php?t=345248
Given two triangles with vertices A1, B1, C1 and A2, B2, C2 respectively. A1A2, B1B2, C1C2 are extended to meet at a point V say. Now, B1C1 and B2C2 are extended to meet at L, A1B1 and A2B2 meet at N and A1C1 and A2C2 meet at M. Prove that L, M and N are concurrent.
Proof (as given in text):
Let A1B1C1 be the reference triangle and V be the unit point (1,1,1). A2 is on the join of A1(1,0,0) and V(1,1,1), so it can be taken as (1+p,1,1). Similarly, the point B2 is given by (1,1+q,1) and C2 by (1,1,1+r).
Now, the line B2C2 is
\left|\stackrel{\stackrel{x}{1}}{1}\stackrel{\stac krel{y}{1+q}}{1}\stackrel{\stackrel{z}{1}}{1+r} \right| = 0.
The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0
i.e. x=0, \frac{y}{q} + \frac{z}{r} = 0
Therefore, L lies on the line \frac{x}{p}+ \frac{y}{q}+ \frac{z}{r} = 0. By symmetry, so do M and N.
Hence proved
From start to finish, I can't get it. Can someone please explain to me what all this means?
For the diagram that goes with the proof, please refer to my topic in the geometry section
http://www.physicsforums.com/showthread.php?t=345248