## 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.

1. anonymous

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

|dw:1437445425378:dw|

3. anonymous

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

I also don't know where to go from here.

5. anonymous

No worries :) Anyone is at least willing to take a look or consider a solution is appreciated

6. mathstudent55

Thanks.

7. anonymous

@ganeshie8

8. jim_thompson5910

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

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

Alright, Im with ya so far :)

11. jim_thompson5910

12. jim_thompson5910

ok I'm going to introduce another variable let y = measure of angle DAC |dw:1437462735373:dw|

13. jim_thompson5910

AI bisects angle CAB for similar reasons why IB bisects angle ABC so angle DAB = angle DAC = y |dw:1437462802570:dw|

14. jim_thompson5910

so arc CD = 2*(angle DAC) = 2y |dw:1437463358939:dw|

15. jim_thompson5910

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

angle BIA = 180 - x - y |dw:1437463483509:dw|

17. jim_thompson5910

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

angle DBI = (angle CBI) + (angle DBC) angle DBI = x + y |dw:1437463668365:dw|

19. jim_thompson5910

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

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

22. anonymous

Yeah, it does, well done. Thanks again

23. jim_thompson5910

you're welcome