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.

B. Conjecture? Prove it! (you may need draw some parallel lines to produce some similar triangles) Also, it probably helps to consider the ratio

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 ABC and triangle DEF is always greater than or equal to 4. When is it equal to 4?

A. If P is centroid then we can know (AF)(BD)(CE) = (FB)(DC)(EA) since AF=FB, BD=DC, and CE=EA.

If ABC is isosceles triangle and P is orthocenter then FBC is congruent to ECB and B is midpoint.

So AF=AE, FB=EC, and BD=DC. Therefore (AF)(BD)(CE) = (FB)(DC)(EA).

To explore this relationship for various triangles and various locations of P through GSP, Click here.

We can see that (AF)(BD)(CE) is always same to (FB)(DC)(EA).

B.Conjecture

is equal to 1

Proof)

Let's draw segments parallel to each side and passing through P.

i.e F"E is parallel to BC, D"F' is parallel to CA, and E"D' is parallel to AB.

I will prove as showing that AF, FB, BD, DC, CE, EA can be expressed by AD, BE, CF, PA, PB, PC, PD, PE, PF using similar triangles.

Since ABD is similar to PD'D and ACD is simailr to PD"D,

AD:PD = AB:PD' = AC:PD" --- (1)

Since BCE is similar to PE'E and BAE is similar to PE"E,

BE:PE = BC:PE' = AB:PE" --- (2)

Since CAF is similar to PF'F and CBF is similar to PF"F,

CF:PF = CA:CF' = CB:PF" --- (3)

Since F"E' is parallel to BC, AF"P is similar to ABD and AE'P is similar to ACD. Thus

AP:AD = F"P:BD = PE': DC --- (4)

Since D"F' is paralle to CA, BD"P is similar to BCE and BF'P is similar to BAE. Thus

BP: BE = D"P:CE = PF':EA --- (5)

Since E"D' is parallel to AB, CE"P is similar to CAF and CD'P is similar to CBF. Thus

CP:CF = E"P:AF = PD':FB --- (6)

By (6), AF = CF/CP*E"P = (CF/CP)*(PE/BE)*AB since (2)

FB = CF/CP*PD' = (CF/CP)*(PD/AD)*AB since (1)

By (5), CE = BE/BP*D"P = (BE/BP)*(PD/AD)*AC since (1)

EA = BE/BP*PF' = (BE/BP)*(PF/CF)*CA since (3)

Let's calculate AF/FB, BD/DC, and CE/EA.

BD/DC = (PF*BE)/(PE*CF)

Therefore,

----- End of proof -----

Can the result be generalized so that point P can be outside the triangle?

We can generalize this fact even though P is outside the triangle. To explore Click here.

C.The ratio of the area of two triangles

We can see that the ratio of the areas of ABC and DEF is always greater than or equal to 4. To explore Click here.

Conjecture:

or euqal to 4

Proof)

At first, I will show that triangle PDE, PEF, PFD can be expressed triangle ABC.

PDE = BDE * (PE/BE) = BCE * (BD/BC) * (PE/BE) = ABC * (CE/AC) * (BD/BC) * (PE/BE)

PEF = CEF * (PF/CF) = AFC * (CE/AC) * (PF/CF) = ABC * (AF/AB) * (CE/AC) * (PF/CF)

PFD = AFD * (PD/AD) = ABD * (AF/AB) * (PD/AD) = ABC * (BD/BC) * (AF/AB) * (PD/AD)

Since DEF = PDE + PEF + PFD,

DEF = ABC * (CE/AC) * (BD/BC) * (PE/BE) + ABC * (AF/AB) * (CE/AC) * (PF/CF) + ABC * (BD/BC) * (AF/AB) * (PD/AD).

I. e. DEF = ABC * ( (CE/AC) * (BD/BC) * (PE/BE) + (AF/AB) * (CE/AC) * (PF/CF) + (BD/BC) * (AF/AB) * (PD/AD))

Therefore,

DEF/ABC = (CE/AC) * (BD/BC) * (PE/BE) + (AF/AB) * (CE/AC) * (PF/CF) + (BD/BC) * (AF/AB) * (PD/AD) --- (*)

We know that

AF = (CF/CP)*(PE/BE)*AB

Since AB = AF + FB and AB = CF/CP*(PE/BE + PD/AD)*AB,

PE/BE + PD/AD = CP/CF = (CF - PF)/CF = 1 - PF/CF

PF/CF + PE/BE + PD/AD = 1

Let AF/AB = a, BD/BC = b, CE/CA = c, PE/BE = x, PF/CF = y, PD/AD = z. Then (*) can be expressed by

DEF/ABC = bcx + acy + baz and x + y + z = 1

I will show that a,b, and c can be expressed by x, y, and z.

Since AF/FB = (PE/BE)/(PD/AD) = x/ z, FB = AB - AF, and a/(1 - a) = x / z,

a = x / (x + z ) and a = x / (1 - y )

Simiary, b = y / (1 - z ) and c = z / (1 - x )

ABC/DEF = xyz/((1-z)(1-x)) + xyz/((1-x)(1-y)) + xyz/((1-y)(1-z)) = xyz(1-y + 1-z + 1-x)/((1-x)(1-y)(1-z)) = 2xyz/((1-x)(1-y)(1-z))

DEF/ABC = 2xyz/(xy + yz + xy - xyz)

We know that

or equal when x=y=z(The relation of Arithmetic mean and Harmonic mean)

Since x + y + z = 1 and xy + yz + zx >= 9xyz,

DEF/ABC <= 2xyz/(9xyz-xyz)

DEF/ABC <= 1/4

Thus ABC/DEF>= 4

Also we know that ABC/DEF=4 when x=y=z=1/3. Therefore

P has to be a centroid for the ratio to be 4

----- End of proof -----

We can notice easily that if P is a centroid then DEF = ABC/4.

Since D, E, and F are midpoints, DEF is similar to ABC with the ratio 1/2.

Therefore the area of DEF is a quarter of ABC