## The Last One!

## Final Project

## 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 ->

## II.b)

## Proof:

## 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)

**Return to Assignment
Matrix:**

Questions? E-mail: **gt0353d@arches.uga.edu**