Hi Vladimir,

Thanks for the Information. I hope the figure comes through OK.

http://geocities.com/bractals/cut-in-half.gif

Analysis of the construction:

Let the line through point P cut the sides AB and AC at points X and Y respectively.

Let c = |AB|, b = |AC|, x = |AX|, and y = |AY|. For the area of DXAY to be one-half the area of DBAC we need xy = bc/2 (*).

In the figure I show two cases: P1 inside the triangle and
P2 outside the triangle.

DXAY ~ DP1R1Y ~ DP2R2Y. Therefore,

|AX| / |AY| = |R1P1| / |R1Y| = |R2P2| / |R2Y|

|AX| / |AY| = |R1P1| / (|AY| - |AR1|) = |R2P2| / (|AR2| - |AY|)

x/y = u1/ (y - v1) = -u2/ (v2 - y) ,

where v = |AR| and |u| = |RP| with u < 0 if P is on the opposite side of line AC from the vertex B.

Therefore, x/y = u/(y - v) (**).

Eliminating y from (*) and (**) we get

2vx2 - bcx + bcu = 0.
Solving for x we get

x = w ± sqrt(w2 - 2uw) , where w = bc/(4v).

By labeling the triangle so that P lies in the yellow area (see the figure), we guarantee that w2 - 2uw > 0 and that X lies between B and C' if we use the plus sign ( I will let you verify this ). Hence,

x = w + sqrt(w2 - 2uw)

By letting t2 = (2|u|)w we have

x = w + sqrt(w2 + t2 ) for P outside the triangle and

x = w + sqrt(w2 - t2 ) for P inside the triangle.

I think you should be able to follow the construction from this. I hope my first attempt with HTML worked.

Bractals