CevaÕs
Theorem Appilcation
By
Shadreck
S Chitsonga
Using CevaÕs Theorem to
prove Concurrency
1. To prove that the medians
of a triangle are
concurrent
In triangle ABC , the points D,E
and F are the midpoints of the sides,BC,CA and AB and CA respectively. This means that (AF) (BD) (CE) = (BF)
(CD) (AE).
It hen follows that 
holds. (i)
According to CevaÕs theorem if
(i) holds the transversals (Cevians) meet at a common point. But we know that 
, 
and 
are the three
angle transversals as well as medial lines. These are concurrent.
The medians are concurrent
2. To prove that bisectors of the angles of the
triangle meet at a point. For the proof go back to the section on proofs.
3.The
perpendiculars from each vertex to the line
of the opposite side of a triangle are
concurrent.
Given: Triangle ABC with CD, AE, AC as
the
perpendiculars to the sides
AB,BC, and AC
respectively.
To Prove: that the perpendiculars are
concurrent
Proof:
In triangles BFP and CEP
It can be shown that triangle BPF is
similar to triangle CPE. Similarly we can
show that triangle
PDC is similar to
triangle AFP and that triangle BPD
is similar to triangle APE We can then
consider the following ratios.
(BF/CE)=(BP/CP)=(FP/EP)
(CP/AP)=(PD/AF)=(DC/FP)
(BP/AP)=(PD/PE)=(BD/AE)
Choosing the ratios we have
(BF/CE) = (BP/CP), (CP/AP)= (DC/FA), (BP/AP)=(PC/PA).
If we arrange these ratios we end up with
(BF) (CD) (AE) = (AF) (CE) (BD)
This implies that
holds. If this
is true then according to
CevaÕs Theorem, the transversals CF, BE and AD pass through a
common
point. But these transversals are perpendiculars (altitudes) to there
respective
sides, therefore they are concurrent.