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.

Select D and E as the bisectors of AB and AC respectively.  Now draw a line from A to G on BC passing thru F.  This is also a bisector of BC.

let the areas of each of these triangles be

BDF=a1

BFG=a2

GFC=a3

CFE=a4

AFE=a5

ADF=a6

now we have

a1=r

a2+a3=q

a4=p

a1/a6=BD/AD=1 => a1=a6

a5/a4=AE/EC=1 => a5=a4

a3/a2=CG/BG=1 => a3=a2

now the area of ABC is simply

a1+a2+a3+a4+a5+a6 and from equations above this is simply

r+q+p+p+r=2p+2r+q

so the area of ABC is 2p+2r+q

Posted by Daniel on 2008-11-30 07:06:11