SOLUTION: A triangle ABC has a median on side AC and AD which meets vertex B and C and the point where both median intersect named O. show that BOC =90˚ +BAC

Algebra ->  Triangles -> SOLUTION: A triangle ABC has a median on side AC and AD which meets vertex B and C and the point where both median intersect named O. show that BOC =90˚ +BAC       Log On


   

Question 174584: A triangle ABC has a median on side AC and AD which meets vertex B and C and the point where both median intersect named O. show that BOC =90˚ +BAC
Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!

Good way to prove your theorem "BOC=90˚+BAC" is via Equilateral Triangle.
Let's see as per the conditions:
---->
RIGHT GRAPH: you see GREEN line from vertex B going to midpoint (median) of AC, and GREEN line from vertex C going to midpoint (median) of AB.
Evaluating:

To PROVE: BOC =90˚ +BAC
As you see the values in the graph:
120%5Eo=90%5Eo%2B60%5Eo
120deg is not equal to 150deg
If we rewrite what to prove:--->BOC=60˚ +BAC, then,
120%5Eo=60%5Eo%2B60%5E0
120%5Eo=120%5Eo, correct
.
*Note: if you increase the angle on BAC, BOC increases too.
Let's see if BAC=90˚:
----->
.Now, we prove: BOC=90˚ +BAC
BOC=90˚ + 90˚
BOC=180˚-----------> NOT TRUE, AS YOU SEE IN THE GRAPH.
.
So if you can double check the condition to prove please.
If you're not satisfied, try to poste it again.
Thank you,
Jojo