anirudh215
Oct13-09, 04:24 AM
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?
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?