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anonymous
 one year ago
In a triangle ΔABC, Let O be the circumcenter, and let I be the incenter. Let the line AI intersect the circumcircle at D. Prove ΔBID is isosceles.
anonymous
 one year ago
In a triangle ΔABC, Let O be the circumcenter, and let I be the incenter. Let the line AI intersect the circumcircle at D. Prove ΔBID is isosceles.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I assume that I just don't know enough about all the centers of a triangle and the related theorems to see the result to this. Either way, don't know what connections to make. All I was able to do was find that a couple of the triangles are similar, but not sure that gets me anywhere. I would assume that the circumcenter has to be involved somehow, but I'm not sure how.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437445425378:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, what you have drawn looks good, similar to what I had drawn up. From there, wasn't sure where to go. All I could do was determine that certain triangles were similar, otherwise no idea.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0I also don't know where to go from here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No worries :) Anyone is at least willing to take a look or consider a solution is appreciated

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2The center I is the center of the incircle. To create the incircle based on the triangle, we find the angle bisectors of triangle ABC so these two angles are congruent (because IB bisects angle ABC) dw:1437461971692:dw angle ABI = angle CBI

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2Let x = angle ABI so angle ABC = 2x the inscribed angle theorem allows us to find arc AC (it's just double that of inscribed angle ABC, so measure of arc AC = 2*2x = 4x) dw:1437462076025:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, Im with ya so far :)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2hmm one sec, I made a bad assumption. Let me rethink

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2ok I'm going to introduce another variable let y = measure of angle DAC dw:1437462735373:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2AI bisects angle CAB for similar reasons why IB bisects angle ABC so angle DAB = angle DAC = y dw:1437462802570:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so arc CD = 2*(angle DAC) = 2y dw:1437463358939:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2using the inscribed angle theorem in reverse lets us find that angle DBC = (arc CD)/2 = 2y/2 = y dw:1437463424601:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2angle BIA = 180  x  y dw:1437463483509:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2the supplement of angle BIA is angle BID = 180  (angle BIA) = 180  (180  x  y) = 180  180 + x + y = x+y angle BID = x+y dw:1437463595501:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2angle DBI = (angle CBI) + (angle DBC) angle DBI = x + y dw:1437463668365:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2At this point, we have angle B = angle I (just focus on triangle BID, nothing else). Both are equal to x+y these are the base angles of the isosceles triangle. You can use further geometry, but at this point, the proof is essentially done. If you have a triangle with congruent base angles, you have an isosceles triangle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, thats all that would be needed. Once you get the two base angles equal or two of the sides equal we're done, so yeah, no need to go further :) That's a lot of running around, not sure I would've come up with something like that. Thanks a lot for coming up with that, understood it all the way through :D

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I'm glad it makes sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, it does, well done. Thanks again

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're welcome
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