## II.a)

For different locations of P, The ratio of (AF*BD*CE ) / (FB*DC*EA) will always be one. To observe an example please click here ->

## Given:

Triangle ABC with Point P interior to ABC

## Prove:

(AF*BD*CE) / (FB*DC*EA) = 1

Let Line L be parallen to line BC. By vertical angles, alternating interior angles we have similarity between the following triangles:

Triangle AEH ~ Triangle CEB

Where the following Properties Hold:

AE/CE = EH/EB = AH/CB

Triangle AFI ~ Triangle BFC

Where the following Properties Hold:

AF/FB = FI/FC = AI/BC

Triangle AIP ~ Triangle DCP

Where the following Properties Hold:

AI/DC = IP/PC = AP/PD

Triangle APH ~ Triangle DPB

Where the following Properties Hold:

AH/BD = PH/PB = AP/PD

We have the following similarity relationships:

AH/BD = PH/PB = AP/PD (Eq.1)

AI/DC = IP/PC = AP/PD (Eq.2)

AF/FB = FI/FC = AI/BC (Eq.3)

AE/CE = EH/EB = AH/CB (Eq.4)

From Eq.4 we get:

AH = (BC*AE)/EC (Eq.5)

From Eq.3 we get:

AI = (BC*AF)/FB (Eq.6)

From Eq.1 and Eq.2 we get:

(AI/DC) = (AH/BD) (Eq.7)

By substituting Eq.5 and Eq6. into Eq.7 =>

[(BC*AF)/(FB*DC)] = [(BC*AE)/(EC*BD)]

Cross multiply:

[(BC*AF)*(EC*BD)] = [(BC*AE)*(FB*DC)]

solve for BC

BC*[AF*EC*BD] = BC*[AE*FB*DC]

BC/BC = [AE*FB*DC]/[AF*EC*BD]

1 = [AE*FB*DC]/[AF*EC*BD]

## II.c)

The ratio of triangle ABC to triangle EFD will equal four only when P is at the centroid of triangle ABC. The ratio will always be four, if you don't believe me, attempt to find a ratio below four for by moving P anywhere in the interior after you click here -> (w javascript) (w/o javascript)