CevaŐs Theorem

I will start my
investigation by considering a triangle ABC and an arbitrary point P inside the
triangle. We will construct line AP, BP, and CP and extend those lines to
intersect the opposite sides at points D, E, and F respectively. I will explore
(AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and locations of P.

Consider the above
construction. The strategy will be to construct a line through P which is
parallel to each of the three sides of ABC. Doing this will allow me to locate
similar triangles and this will help me to relate their sides.

By constructing a line through P and parallel to AB, I can
determine congruent angles which leads to similar triangles.

By the corresponding angle
theorem, angle AFC is congruent to angle HPC and angle FAC is congruent to
angle PHC. This implies that triangles AFC and HPC are similar or AF/HP=FC/PC,
by AA Similarity Theorem.

By the same argument I can
say that triangles BFC and JPC are similar or BF/JP=FC/PC. By the above I could
also write AF/BF = HP/JP.

I can also see that triangles
BAE and HPE are similar, then HP/AB = PE/BE.

Furthermore, I see that
triangle BAD is similar to triangle JPD, then JP/BA=PD/AD.

So, AF/BF = HP/JP =
(HP/AB)/(JP/AB) = (PE/BE)/(PD/AD).

Now, letŐs construct another
parallel line through P parallel to BC.

As I did before, I will
locate similar triangles.

Triangle ABD is similar to
triangle AIP and PAK is similar to DAC resulting in BD/IP = AD/AP and DC/PK =
AD/AP. This, then implies BD/DC = IP/PK and PE/BE = PK/BC giving me FP/FC =
IP/BC.

Therefore, BD/DC =
(PJ/BC)/(PK/BC) = (FP/FC)/(PE/BE).

Finally, letŐs consider a
line through P parallel to AC.

Now, letŐs locate similar
triangles.

Triangle BEC is similar to
triangle BPM, so EC/PM = BE/BP.

Triangle ABE is similar to
triangle LBP, so AE/LP = BE/BP which implies that EC/PM = AE/LP.

Triangle CDA similar to
triangle DPM, so PM/AC = PD/AD.

Triangle AFC is similar to
triangle FLP, so LP/AC = FP/FC which implies that EC/AE = PM/LP =
(PM/AC)/(LP/AC)=(PD/AD)/(FP/FC)

At last, letŐs put this all
together.

**(AF/FB)*(BD/DC)*(EC/EA) =
(PE/BE)/(PD/AD) * (FP/FC)/ (PE/BE) *(PD/AD)/(FP/FC) = 1**

** **

**So this proves that (AF/FB)*(BD/DC)*(EC/EA)
= 1.**

** **

** **

** **

** **

**Next, letŐs use CevaŐs
Theorem to show the concurrency of altitudes of a triangle.**

** **

I know that AD is the
altitude of BC, BE is the altitude of AC, and CF is the altitude of AB.

By the Angle-Angle Similarity
Theorem, I know that triangle AFD is similar to triangle CDP, triangle BFP is
similar to CEP and triangle BDP is similar to triangle AEP.

Thus, (AF/CD)=(AP/CP)=(FP/DP)
, (BF/CE) = (BP/CP) = (FP/EP) and (BD/AE)=(BP/AP)=(DP/EP).

Which implies that
(CP)(FP)=(AP)(DP), (CP)(FP)=(BP)(EP), and (BP)(EP)=(AP)(DP).

So,
(AP/CP)=(FP/DP)=(BP/CP)=(FP/EP)=(BP/AP)=(DP)(EP), therefore
(AF/CD)=(BF/CE)=(BD/AE).

Finally, I can conclude that
(CD/BD)*(BF/AF)*(AE/CE)=1.

Therefore the converse
of CevaŐs Theorem confirms that
the altitudes are concurrent.

Next, letŐs use CevaŐs
Theorem to show the concurrency of angle bisectors of a triangle.

Considering the following
figure, AD is the angle bisector of angle A, FC is the angle bisector of angle
C, and BE is the angle bisector of angle B. By the Angle Bisector Theorem I
know that (BD/CD) = (AB/AC) and similarly (AE/CE) = (AB/BC) and (AF/BF) =
(AC/BC).

This implies that AB = ((BD)(AC))/CD and that AB = ((AE)(BC))/CE.

So, by substitution, I know
that ((BD)(AC))/CD = ((AE)(BC))/CE

Cross multiplication gives us
(BD)(AC)(CE)=(CD)(AE)(BC) which is ((BD)(AC)(CE))/((CD)(AE)(BC)) = 1.

From above, I also know that
AC = ((AF)(BC))/BF. Now by substituting
AC = ((AF)(BC))/BF into ((BD)(AC)(CE))/((CD)(AE)(BC)) = 1, I receive
((BD)(CE)(AF))/(CD)(AE)(BF)) = 1. So CevaŐs Theorem confirms the concurrency of
angle bisectors.

Now letŐs use CevaŐs Theorem
to show the concurrency of medians of a triangle.

Consider the following
construction, AD bisects BC, BE bisects AC and CF bisects AB.

Since AF=BF, BD=CD and AE=CE,
I can say that (AF/BF)=(BD/CD)=(AE/CE)=1, so ((AF)(BD)(CE))/((BF)(CD)(AE))=1.
Therefore, CevaŐs Theorem confirms that the altitudes of a triangle are
concurrent.

Lastly, what happens when P
is outside of the triangle? After viewing a GSP sketch it appears that CevaŐs
Theorem still holds when P is outside the triangle.

Click here
for a GSP file to further explore this case.