Let Γ be the circumcircle of ΔABC and let D be the midpoint of the arc BC. Prove that the incenter I of ΔABC lies on the circle centered at D that goes through B.

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Let Γ be the circumcircle of ΔABC and let D be the midpoint of the arc BC. Prove that the incenter I of ΔABC lies on the circle centered at D that goes through B.

Mathematics
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Do you really mean "lies on the curcumference of circle centered at D" ?
Well, lies on the boundary, edge, I guess |dw:1437879668554:dw|
Because after playing a bit, im getting that " the incenter lies interior to the circle centered at D" Not completely sure though.. still working..

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I checked it on a program and it seems to work. Ill try and get the screenshot.
Basically we want to show \(DB=DI=DC\) |dw:1437879935160:dw|
Wish I knew how to upload it faster, lol http://i201.photobucket.com/albums/aa120/ApocalypticDeity/6_zpsq2dhiksj.png
Interesting, do we know if AID lie on a straight line from some theorem or somehting ? |dw:1437880349134:dw|
I dont know actually. That picture I posted was the original picture plus me trying out a few things. It looked like they may be collinear, so I tried it and they were. No explanation that I could give, though.
btw i need to start over, i made a stupid mistake in my earlier attempt
|dw:1437880661407:dw|
Well, we want to show that DI would be the same as DB and DC. I assumed that was the way to try and tackle it, see if I can show that DI is a radii. I didn't find any similarity with the triangles where the two circles overlap. The fact A, I, and D were collinear was really interesting, but even if it was useful Id still have to state why theyre collinear, which I dont know.
|dw:1437881020931:dw|
|dw:1437881129726:dw|
|dw:1437881183330:dw|
equaliteral triangle so we are gone i think
done
oh wait that assumes the incenter of the triangle is the center of the circumcircle too which is not true is it?
I is incenter dear, not circumcenter
Yeah, I is incenter and I would mark O as circumcenter. The picture I posted has the circumcenter on it for reference if you think it may be helpful.
|dw:1437881398303:dw|
No idea if the circumcenter would help. I don't think there's a way to set up an inscribed angle theorem. I don't know if there is a more proper name for the theorem, but I was also trying to see if I could use power of the point. I know we can relate area to the inradius or use extended law of sines to have a relation to the radius of the circumcircle, but I don't think any of that is useful, lol. Just random connections and theorems and such.
Gotcha!
cyclic quadrilateral does the job !
I was considering that also, but is it cyclic?
|dw:1437881873780:dw|
Oh, different quadrilateral than what I was looking at, lol.
|dw:1437881954866:dw|
That angle is arbitrarily inscribed on that circle or is it in a specific spot?
bottom is an arbitrarily inscribed angle top is fixed, incenter
ok got it
Okay, that definitely is cyclic then :P And yeah, that is just the circumcircle, I is not the center of it. This was the picture from my program, dan. The actual center of the top circle I have marked as O. It's kind of hidden, though http://i201.photobucket.com/albums/aa120/ApocalypticDeity/6_zpsq2dhiksj.png
That proves BIC lie on the same circle since 3 points uniquely determine a circle, it must be the case that DB=DI=DC
Are we just observing that the one angle subtends an arc have the measure that angle A does?
Hmm didn't get you
Well, in the labeled picture we have angles marked as A/2, B/2, C/2. I was trying to make sure how we knew those were half the measures of angles A, B, and C.
Ahh okay, let me put the complete proof, it shouldn't be long
B/2 and C/2 are from the angle bisectors, but A/2 wasnt as direct. I assumed it was the same logic, but wanted to confirm.
\(\angle DBC\) is the inscribed angle of arc \(DC\), by inscribed angle formula we have \(\angle DBC = \dfrac{\text{arc}DC}{2} = \dfrac{A}{2}\) |dw:1437882784370:dw|
Also, it is important that we don't assume \(I\) is on the circle in the start
Right :)
so that in the end we can use cyclic thingy to conclude that \(I\) lies on the circle that passes through remaining 3 vertices of the quadrilateral
let me know if it is not clear enough, il try writing the complete proof in one reply..
Will do.
wonder if there is an easy way to prove AID are colliner |dw:1437884264926:dw|
Whoa, how'd you make that?
familiar with geogebra ?
That's what I used to make my picture, but today is the first day I've ever messed with it, so no idea of its capabilities yet
its pretty good and very easy to learn
let me upload the file to geogebratube so that you can play with it
Alright, cool, thanks :)
Here it is http://tube.geogebra.org/m/1437801
that is very much like a geogebra session, you can do everything that you can do with geogebra on ur computer.. all commands work just fine
Gotcha. I have to get familiar with Geometer's Sketchpad really soon, too. It's the required software for my course when we get to that part of the class. But yeah, I think I found a way to show the quadrilateal is cyclic, but not sure it's the best way.
if psble pls share, id like to see
Since \(BI\) and \(CI\) are the angle bisectors of B and C, \(\angle IBC = B/2\) and \(\angle ICB = C/2\). The sum of the angles of a triangle sum to 180 degrees, thus \(\angle BIC = 180° - \frac{B + C}{2}\). \(OB\) and \(OC\) are both radii and thus have the same measure. This implies \(\angle OCB = \angle OBC\). By the inscribed angle theorem, \(\angle OCB = \angle OCB = A/2\). Furthermore, \(\angle BOC = 180° - A\). Draw the lines BE and CE where E is an arbitrary point on the circle of radius D and on the arc counterclockwise from B to C (Or however you would phrase that). By inscribed angle theorem, \(\angle BEC = \angle BOC/2 = 90° - A/2\). From here, \(\angle BEC + \angle BIC = 180° - \frac{B + C}{2} + 90° - A/2 = 270° - \frac{A + B + C}{2} = 180°\). Since opposite angles of a quadrilateral sum to 180°, quadrilateral IBEC is cyclic. Bleh, seems chunky, but idk, lol.
looks okay to me!

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