At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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.
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.
I also don't know where to go from here.
No worries :) Anyone is at least willing to take a look or consider a solution is appreciated
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
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|
Alright, Im with ya so far :)
hmm one sec, I made a bad assumption. Let me rethink
ok I'm going to introduce another variable let y = measure of angle DAC |dw:1437462735373:dw|
AI bisects angle CAB for similar reasons why IB bisects angle ABC so angle DAB = angle DAC = y |dw:1437462802570:dw|
so arc CD = 2*(angle DAC) = 2y |dw:1437463358939:dw|
using the inscribed angle theorem in reverse lets us find that angle DBC = (arc CD)/2 = 2y/2 = y |dw:1437463424601:dw|
angle BIA = 180 - x - y |dw:1437463483509:dw|
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|
angle DBI = (angle CBI) + (angle DBC) angle DBI = x + y |dw:1437463668365:dw|
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
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
I'm glad it makes sense
Yeah, it does, well done. Thanks again