Bouncing Barney
& CevaÕs Theorem
By
Princess Browne
Barney
is in the triangular room shown below. He walks from a point on BC parallel to
AC. When he reaches AB, he turns and walks parallel to BC. When he reaches AC,
he turns and walks parallel to AB.
 Prove
that Barney will eventually return to his starting point.
 How
many times will Barney reach a wall before returning to his starting point?

Explore and discuss for various starting points on line BC, including points
exterior to segment BC. Discuss and prove any mathematical conjectures you
find.
In
the exploration, we will investigate different starting points for Barney to
determine how many times he will bounce off a wall, before returning to his
starting point.
It appears that Barney will return to his starting
point after bouncing off each wall once only if he starts from a midpoint of
one of the segments or if he starts from one of the vertices of triangle ABC. We
know that when a line is parallel to one side of a triangle and passes through
the other two sides, then it divides these two sides into equal segments. So if
Barney starts at the midpoint of segment BC, he will hit the midpoint of
segments AB and AC before returning to his starting point.
By definition of the Auxiliary or Medial triangle: Joining
the midpoints of the sides of triangle ABC will form DDEF. Also from the properties of midsegment of a
triangle we know that by joining the midpoint of two sides of a triangle its
length is half the length of the third side of the triangle.
F is
the midpoint of segment AC and E is the midpoint of segment AB. Midsegment EF
is parallel to BC, EF = 1/2AC
Next, we want to show that if Barney starts anywhere
other than the vertices and the midpoint he will bounce off each segments twice
before returning to his starting point.
From
the pictures above, we can make the following conjectures: The perimeter of the
triangle ABC is equal to the sum of the bounces.
To
prove that triangle ABC is equal to the perimeter of DIGHIH we will show the
following: Since BH = DI and GF = AH, then DI + GF = AB. Also, CI = GH and CF =
DE, which means that HG + DE = AC. Finally, BG = EF and CG = HI, which also
means that EF + HI = BC.
What
will happen if Barney starts outside of triangle ABC? He will eventually return
to his starting point, no matter if the point is inside or outside triangle
ABC.
Because
BarneyÕs path will always be parallel to the three sides of triangle ABC, we
know the following: HIBC EF, DE ACHG, and FGABDI which means angle
IDC = angle ABC because they are corresponding angles of parallel lines AB and
DI.
CevaÕs Theorem
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.
Consider
any triangle ABC, if the points
D,E, and F lie on segments BC, AC,
and AB of the triangle, then the lines AD, BE, and CF are concurrent if and only if the product of
the ratio (AF/FB) . (BD/DC). (CE/EA) = 1
AD,
BE, and CF are concurrent if and only if
Sin(ÐBAD)/Sin(ÐCAD)* sin(ÐACF)/sin(ÐBCF)* sin(ÐCBE)/( ÐABE) = 1
What
if AD, BE, and CF interest at P. Because DBPD and DCPD have the same height we will have
¨DBPD¨/¨DCOD¨ = BD/DC similarly ¨DBAD¨/¨DCAD¨= BD/DC
BD/DC
= ¨DBAD¨/¨DCAD¨ ¨DBPD¨/¨DCPD¨= ¨DABP¨ /¨DCAP¨
and similarly
CE/EA
= ¨DBCP¨/¨DABP¨and AF/FB =¨DCAP¨/¨DBCP
Finally,
LetÕs use CevaÕs theorem to prove that that the altitude of a triangle are
concurrent.
We
know that D
CAF is similar to DBAE from the following: (CA/BA) = (CF/BE) = (AF/AE)
DABD is similar to DCBF by the following: (AB/CB) =(AD/CF) = (BD/BF)
DACD is similar to DBCE by the following: (AC/BC) = (AD/BE) =(CD/CE)
Which yields
(AF/FB)(BD/DC)(EC/EA) = (AF/EA)(BD/DB)(CE/DC) = (CF/BE)(AD/CF)(BE/AD) =
1
CevaÕs theorem holds even if p is outside of the triangle.
CevaÕs theorem can be use to prove angle bisectors, altitude, and point of concurrency for triangle medians.