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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.

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  1. anonymous
    • one year ago
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    I 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.

  2. mathstudent55
    • one year ago
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    |dw:1437445425378:dw|

  3. anonymous
    • one year ago
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    Right, 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.

  4. mathstudent55
    • one year ago
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    I also don't know where to go from here.

  5. anonymous
    • one year ago
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    No worries :) Anyone is at least willing to take a look or consider a solution is appreciated

  6. mathstudent55
    • one year ago
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    Thanks.

  7. anonymous
    • one year ago
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    @ganeshie8

  8. jim_thompson5910
    • one year ago
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    The 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

  9. jim_thompson5910
    • one year ago
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    Let 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|

  10. anonymous
    • one year ago
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    Alright, Im with ya so far :)

  11. jim_thompson5910
    • one year ago
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    hmm one sec, I made a bad assumption. Let me rethink

  12. jim_thompson5910
    • one year ago
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    ok I'm going to introduce another variable let y = measure of angle DAC |dw:1437462735373:dw|

  13. jim_thompson5910
    • one year ago
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    AI bisects angle CAB for similar reasons why IB bisects angle ABC so angle DAB = angle DAC = y |dw:1437462802570:dw|

  14. jim_thompson5910
    • one year ago
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    so arc CD = 2*(angle DAC) = 2y |dw:1437463358939:dw|

  15. jim_thompson5910
    • one year ago
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    using the inscribed angle theorem in reverse lets us find that angle DBC = (arc CD)/2 = 2y/2 = y |dw:1437463424601:dw|

  16. jim_thompson5910
    • one year ago
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    angle BIA = 180 - x - y |dw:1437463483509:dw|

  17. jim_thompson5910
    • one year ago
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    the 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|

  18. jim_thompson5910
    • one year ago
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    angle DBI = (angle CBI) + (angle DBC) angle DBI = x + y |dw:1437463668365:dw|

  19. jim_thompson5910
    • one year ago
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    At 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

  20. anonymous
    • one year ago
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    Yeah, 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

  21. jim_thompson5910
    • one year ago
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    I'm glad it makes sense

  22. anonymous
    • one year ago
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    Yeah, it does, well done. Thanks again

  23. jim_thompson5910
    • one year ago
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    you're welcome

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