& CevaÕs Theorem
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 mid-segment 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. Mid-segment 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: HI||BC ||EF, DE ||AC||HG, and FG||AB||DI which means angle IDC = angle ABC because they are corresponding angles of parallel lines AB and DI.
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¨
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.