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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

[tex]\left|\stackrel{\stackrel{x}{1}}{1}\stackrel{\stackrel{y}{1+q}}{1}\stackrel{\stackrel{z}{1}}{1+r} \right|[/tex] = 0.

The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0

i.e. x=0, [tex]\frac{y}{q} + \frac{z}{r}[/tex] = 0

Therefore, L lies on the line [tex]\frac{x}{p}+ \frac{y}{q}+ \frac{z}{r}[/tex] = 0. By symmetry, so do M and N.

Hence proved

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

[tex]\left|\stackrel{\stackrel{x}{1}}{1}\stackrel{\stackrel{y}{1+q}}{1}\stackrel{\stackrel{z}{1}}{1+r} \right|[/tex] = 0.

The point L is given by x = 0, y{1-(1+r)} + z{1 - (1+q)} = 0

i.e. x=0, [tex]\frac{y}{q} + \frac{z}{r}[/tex] = 0

Therefore, L lies on the line [tex]\frac{x}{p}+ \frac{y}{q}+ \frac{z}{r}[/tex] = 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

https://www.physicsforums.com/showthread.php?t=345248