this is very hard for me... you mean ruler and compass construction?>??

Let it be triangle ABC, and WLOG let P lie on AB somewhere. Note that the line dividing it into halves must pass through one of the other sides. Extend AC, and join P to C. Now from B, construct a line parellel to PC, and call where it hits the extended AC point D. Bisect AD. If the point of bicestion lies between A and C, join P to this point, and we have our line. If it lies outside point C and the line AC, repeat the above instruction, but swap point B with point C, ie come from, A parellel to PC to where it meets BC etc etc. Since the line must pass over one of these lines, one of the bisectors must be on the inside of the triangle. Im in a hurry, see if you can prove it (its not hard) but ill be back. PM for help

Assume without loss of generality that p lies on side bc and is closer to b than c. (If it is the midpiont than just connect it to A). Draw the midpoint of line BC and call it D. if we can find a point S on side AC such that traingles SPA and APD have the same area then we are done. To do this draw the height of triangle APD with base AP and draw a line perpendicular to it at point D and where that line intersects line AC is S.