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anonymous
 one year ago
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.
anonymous
 one year ago
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Do you really mean "lies on the curcumference of circle centered at D" ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, lies on the boundary, edge, I guess dw:1437879668554:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Because after playing a bit, im getting that " the incenter lies interior to the circle centered at D" Not completely sure though.. still working..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I checked it on a program and it seems to work. Ill try and get the screenshot.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Basically we want to show \(DB=DI=DC\) dw:1437879935160:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wish I knew how to upload it faster, lol http://i201.photobucket.com/albums/aa120/ApocalypticDeity/6_zpsq2dhiksj.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Interesting, do we know if AID lie on a straight line from some theorem or somehting ? dw:1437880349134:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1btw i need to start over, i made a stupid mistake in my earlier attempt

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437880661407:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1equaliteral triangle so we are gone i think

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh wait that assumes the incenter of the triangle is the center of the circumcircle too which is not true is it?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I is incenter dear, not circumcenter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1cyclic quadrilateral does the job !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was considering that also, but is it cyclic?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437881873780:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, different quadrilateral than what I was looking at, lol.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437881954866:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That angle is arbitrarily inscribed on that circle or is it in a specific spot?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1bottom is an arbitrarily inscribed angle top is fixed, incenter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1That proves BIC lie on the same circle since 3 points uniquely determine a circle, it must be the case that DB=DI=DC

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are we just observing that the one angle subtends an arc have the measure that angle A does?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh okay, let me put the complete proof, it shouldn't be long

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0B/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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(\angle DBC\) 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Also, it is important that we don't assume \(I\) is on the circle in the start

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1so 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let me know if it is not clear enough, il try writing the complete proof in one reply..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1wonder if there is an easy way to prove AID are colliner dw:1437884264926:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoa, how'd you make that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1familiar with geogebra ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That'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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1its pretty good and very easy to learn

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let me upload the file to geogebratube so that you can play with it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, cool, thanks :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Here it is http://tube.geogebra.org/m/1437801

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Gotcha. 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1if psble pls share, id like to see

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since \(BI\) 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.
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