**This time the construction
has to be done more carefully. For better investigation, let the
two triangle not be separetely constructed as earlier. Now construct
triangle ABC as shown in the figure below. Let CE, BD and AF be
the medians of triangle ABC. Using median CE and point E construct
a ray parallel to DB and also construct another ray from C, parallel
to CB to meet the other ray at point G. Using points CEG construct
segments to form triangle GEC as shown below. Click here to see gsp file**

**Note that as per the construct,
segment GE is paralle to DB, and AF is parallel to CG, taking
EC as the transversal line, angle GEC is corresponding to DOC,
hence equal. Let angle GCE be X degrees, angle DOC will be supplimentary
to BOC hence angle BOC will be (180 - x)º. With the same
arguement, angle EOB will Xº.**

**Before I go further to proof
this conjecture, it should be noted that medians have two main
properties**

**Point of intersection of the two medians CE and BD devides each of them in the ratio of 2:1. (i)****Medians devide area of triangle in two halfs, for example median EC devide area ABC into two equal areas.(ii)**

**With that in mind, let us
consider triangle ABC and median EC which devide it in two equal
areas. Therefore area EBC is = .**

**I will find the area of triangle EBC which
is area of triangle EOB + area of triangle BOC. To find area of
a triangle, I will use the formula where
a and are sides that enclose the angle.**

**Area of triangle **

**Area of triangle **

**Now let EC = y and DB = t, then using property
(i) then OC = 2/3EC = 2/3y, where as OB = 2/3DB = 2/3t**

**therefore area of triangle**

**and area of triangle**