12+2

The point at which they all intersect creates a ratio on each median. The ratio being 2/1. Example: if we create a triangle in which A,B,C are the angles, and D,E,F are the points of median, and O is the point at which all medians intersect then from BD and AF: AO/OF=BO/OD=2/1 then AO=2OF. Create a second triangle with the center point being Q. AQ/QF=CQ/QE=2/1 then AQ=2QF then Q=O. The point is always the same for every triangle.
 * __Medians__** - What can we say about them?

Triangle ABC is similar to triangle A'B'C'. Altitudes of triangle ABC are perpendicular bisectors of triangle A'B'C' then perpendicular bisectors are concurrent then altitudes are concurrent. AIA is needed for similarity. Question: Is the circumcenter of a triangle always orthocenter of the midpoint triangle (midtriangle)? Other things we discovered the ratio of the sides are 2/1 but the area is 4/1.
 * __Altitudes__** - What can we say about them?

In a triangle ABC, three lines AD, BE and CF intersect at a single point K if and only if AF/FB*BD/DC*CE/EA=1 Proof: from Triangle AFH is similar to triangle BFC, AF/FB=AH/CB. triangles GAE similar to triangle BCE, CE/AE=BC/AG. triangle BDK similar to triangle GAK, BD/DK=GA/AK then AG/BD=AK/DK. triangle KAH similar to triangle KDC, AH/DC=AK/DK with this you can then say AG/BD=AH/DC. AF/FB*BD/DC*CE/EA=AH/CB*AG/AH*BC/AG=1 then AF/FB*BD/DC*CE/EA =1 K point of intersection of BE and CF extend AK so that line AK line BC= D' using direction we just proved AF/FB*BD'/D'C*CE/EA=1=AF/FB*BD/DC/CE/EA 1+BD'/D'C=BD/DC+1 (D'C+BD')/D'C=(DB+DC)/DC BC/D'C=BC/DC D=D'
 * __Ceva's Theorem__**