Question

If triangle ABC, CD, and AE are medians crossing at O
Prove: Triangle ODE is Similar to Triangle OCA
Drawing Below vvvvv

Answer 1

Okay, let's go about this the angle-angle method.


You're looking for triangle ODE ~ triangle OCA. We can prove by the vertical angles congruence theorem that angle EOD is congruent to angle AOC, as they are clearly vertical angles. One angle down already, one to go.


Now for the second. So, the definition of a median says that a median runs from the midpoint of a side to the opposite vertex. That means E and D are midpoints, and ED is a therefore midsegment. (NOT a median!)

The midsegment theorem says that midsegments are parallel to (and 1/2 the length of, but that doesn't matter here) the opposite side. So, ED || AC.

Now we can use CD as a transversal between the two parallel lines ED and AC. By the alternate interior angles theorem, angle ODE is congruent to angle OCA. There you have it, the second angle!


Because <EOD=<AOC and <ODE=<OCA, triangle ODE ~ triangle OCA.