Question 174584
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Good way to prove your theorem "BOC=90&#730;+BAC" is via Equilateral Triangle.
Let's see as per the conditions:
{{{drawing(400,400,-7,7,-1,7,graph(400,400,-7,7,-1,7),triangle(-6,0,0,6,6,0),locate(-6.2,.4,A),locate(6.2,.4,C),locate(.2,6.2,B),red(locate(-5.2,.7,60^o)),red(locate(0.2,5.5,60^o)),red(locate(4.6,.7,60^o)))}}}---->{{{drawing(400,400,-7,7,-1,7,graph(400,400,-7,7,-1,7),triangle(-6,0,0,6,6,0),locate(-6.2,.4,A),locate(6.2,.4,C),locate(.2,6.2,B),green(line(-3,3,6,0)),green(line(0,6,0,0)),red(locate(-.4,2,O)))}}}
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:
{{{drawing(400,400,-7,7,-1,7,triangle(-6,0,0,6,6,0),locate(-6.2,.4,A),locate(6.2,.4,C),locate(.2,6.2,B),green(line(-3,3,6,0)),green(line(0,6,0,0)),locate(-.4,2,O),red(locate(-5.2,.7,60^o)),red(locate(-1,5,30^o)),red(locate(.4,5,30^o)),red(locate(4.6,1,30^o)),red(locate(4,.5,30^o)),red(locate(0,2.5,120^o)))}}}
To PROVE: BOC =90&#730; +BAC 
As you see the values in the graph:
{{{120^o=90^o+60^o}}}
120deg is not equal to 150deg
If we rewrite what to prove:--->BOC=60&#730; +BAC, then,
{{{120^o=60^o+60^0}}}
{{{120^o=120^o}}}, correct
.
*Note: if you increase the angle on BAC, BOC increases too.
Let's see if BAC=90&#730:
{{{drawing(400,400,-1,13,-1,13,graph(400,400,-1,13,-1,13),triangle(0,0,0,12,12,0),green(line(0,1,1,1)),green(line(1,1,1,0)),blue(locate(-.8,0.8,A)),blue(locate(0.5,12.2,B)),blue(locate(12.2,0.8,C)))}}}----->{{{drawing(400,400,-1,13,-1,13,graph(400,400,-1,13,-1,13),triangle(0,0,0,12,12,0),green(line(0,1,1,1)),green(line(1,1,1,0)),blue(locate(-.8,0.8,A)),blue(locate(0.5,12.2,B)),blue(locate(12.2,0.8,C)),green(line(12,0,0,6)),green(line(0,12,6,0)),blue(locate(3.6,4,O)))}}}
.Now, we prove: BOC=90&#730; +BAC
BOC=90&#730; + 90&#730
BOC=180&#730-----------> 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</pre>