**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.

Conjecture ?

The ratio

Let angle AFP = x, angle BDP = y, angle CEP = z , angle DPC = p, angle BPF = q , angle APE = r.

Let's consider the area of small triangles.

BFP = 1/2(BF)(FP)sinx, BFP = 1/2(BP)(FP)sinq---------------------2

BPD = 1/2(BD)(DP)siny, BPD = 1/2(BP)(DP)sinr-------------------3

CPD = 1/2(CD)(DP)siny, BPD = 1/2(CP)(DP)sinp------------------4

PCE = 1/2(PE)(EC)sinz, PCE = 1/2(PE)(PC)sinq--------------------5

APE = 1/2(PE)(EA)sinz, PCE = 1/2(PE)(AP)sinr---------------------6

BD/BC = (BP/CP)sinr/sinp

CE/AE = (CP/AP)sinq/sinr

**B.** Can the result be generalized (using lines rather than segments
to construct ABC) so that point P can be outside the triangle? Show a working
GSP sketch.

**C. **Show that when P is inside triangle ABC, the ratio of the
areas of triangle DEF is always grater than or equal to 4. When is it equal
to 4?

When P is a centroid of triangle ABC, it is equal to 4.

Triangle ABC is similar to triangle DEF, and AB: DE = BC: EF = CA : FD = 2 : 1.

So the ratio of areas is equal to 4. Q.E.D.

Click **here**
to see a GSP file.