Question

Let P be an interior point of the ∆ABC. Assume that AP, BP , CP meet the opposite sides BC, CA, AB at D , E and F respectively. Prove that

AF/FB + AE/EC = AP/PD

Answer 1

Given: Let P be an interior point of the ∆ABC. Assume that AP, BP , CP meet the opposite sides BC, CA, AB at D , E and F respectively.

To find: Prove that  AF/FB + AE/EC = AP/PD

Solution:

• Now we know that A, B, C, P are co-planar.

• So there exists four scalars x,y,z,w, not all zero simultaneously such that:

                   xA + yB + zC + wP = 0

• Now here x + y + z + w = 0

• Now xA + wP / x+w = yB + zC / y+z

                   So AP/PD = -w/x - 1

• Similarly:

                    xA + yB / x+y = zC + wP / z+w

• So AF/FB = y/x

• Similarly: AE/EC = z/x

• Now x + y + z + w = 0 is true for  -w/x - 1 =  y/x +  z/x

• Hence proved.

Answer:

            So we proved that  -w/x - 1 =  y/x +  z/x