Let ABC be any triangle. Let D be a point on side AB and let E be a point on side AC. Draw lines CD and BE and call their intersection F. Triangle ABC is then divided into three smaller triangles BDF, CEF, BCF and a quadrilateral ADFE.

Let the area of BDF equal r, the area of CEF equal q, and the area of BCF equal p. Express the area of the whole triangle ABC in terms of p, q, and r.

Draw AF and let area ADF = M and area AEF = N.

Areas of triangles with equal altitudes are proportional to their bases, so for BCD, p/r = CF/FD and for BCE, p/q = BF/FE.

Also for ABE, (r + M)/N = BF/FE and for ACD, (N + q)/M = CF/FD.

This gives 2 equations in M and N and after solving you find:

M = qr(p + r)/(p^2 -qr)

N = qr(p + q)/(p^2 - qr)

Posted by xdog on 2008-11-30 23:05:24