Final Assignment

__Part Two__

A.Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively. Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.

From the above four triangles, we can see that (AF)(BD)(EC) = (FB)(DC)(EA).

**B**. **Prove the above.**

We can get the following figure by constructing lines through B and C parallel to AD and extending their intersections through CF and BE. The three segments BX, AD, and CY are parallel.

Because of equal vertical angles and equal interior angles,
triangle **AFP ~ BFX**, and triangle **CEY ~ AEP**. From
this similarity, we can get the following ratios:

**AF/BF = AP/BX** and **CE/AE = CY/AP**

Because of the common angles and equal corresponding angles, triangle BDP ~ BCY and triangle BCX ~ DCP . And the following ratios:

**BD/BC = DP/CY** and **BC/DC = BX/DP**.

Now let's multiply the left and the right sides of these equations to get

(AF/BF)(BD/BC)(BC/DC)(CE/AE)=(AP/BX)(DP/CY)(BX/DP)(CY/AP) and simplifying we get,

(AF/BF)(BD/DC)(CE/AE) = 1 and this means that (AF)(BD)(CE) = (BF)(DC)(AE) as required.

And this ratio will also hold true even when the point P is outside of triangle ABC as shown below.

The above two figures show that the ratios are greater than or equal to 4.

The ratio is exactly 4 when P is the centroid of the triangle as shown below.

Part ThreeCourse Evaluation sent to

PWILSON@coe.uga.edu.