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

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- ganeshie8

Do you really mean "lies on the curcumference of circle centered at D" ?

- anonymous

Well, lies on the boundary, edge, I guess
|dw:1437879668554:dw|

- ganeshie8

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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## More answers

- anonymous

I checked it on a program and it seems to work. Ill try and get the screenshot.

- ganeshie8

Basically we want to show \(DB=DI=DC\)
|dw:1437879935160:dw|

- anonymous

Wish I knew how to upload it faster, lol
http://i201.photobucket.com/albums/aa120/ApocalypticDeity/6_zpsq2dhiksj.png

- ganeshie8

Interesting, do we know if AID lie on a straight line from some theorem or somehting ?
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- anonymous

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.

- ganeshie8

btw i need to start over, i made a stupid mistake in my earlier attempt

- ganeshie8

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- anonymous

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.

- ganeshie8

@dan815

- dan815

|dw:1437881020931:dw|

- dan815

|dw:1437881129726:dw|

- dan815

|dw:1437881183330:dw|

- dan815

equaliteral triangle so we are gone i think

- dan815

done

- dan815

oh wait that assumes the incenter of the triangle is the center of the circumcircle too which is not true is it?

- ganeshie8

I is incenter dear, not circumcenter

- anonymous

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.

- dan815

|dw:1437881398303:dw|

- anonymous

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.

- ganeshie8

Gotcha!

- ganeshie8

cyclic quadrilateral does the job !

- anonymous

I was considering that also, but is it cyclic?

- ganeshie8

|dw:1437881873780:dw|

- anonymous

Oh, different quadrilateral than what I was looking at, lol.

- ganeshie8

|dw:1437881954866:dw|

- anonymous

That angle is arbitrarily inscribed on that circle or is it in a specific spot?

- ganeshie8

bottom is an arbitrarily inscribed angle
top is fixed, incenter

- dan815

ok got it

- anonymous

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

- ganeshie8

That proves BIC lie on the same circle
since 3 points uniquely determine a circle, it must be the case that DB=DI=DC

- anonymous

Are we just observing that the one angle subtends an arc have the measure that angle A does?

- ganeshie8

Hmm didn't get you

- anonymous

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.

- ganeshie8

Ahh okay, let me put the complete proof, it shouldn't be long

- anonymous

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.

- ganeshie8

\(\angle DBC\) 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

Also, it is important that we don't assume \(I\) is on the circle in the start

- anonymous

Right :)

- ganeshie8

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

- ganeshie8

let me know if it is not clear enough, il try writing the complete proof in one reply..

- anonymous

Will do.

- ganeshie8

wonder if there is an easy way to prove AID are colliner
|dw:1437884264926:dw|

- anonymous

Whoa, how'd you make that?

- ganeshie8

familiar with geogebra ?

- anonymous

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

- ganeshie8

its pretty good and very easy to learn

- ganeshie8

let me upload the file to geogebratube so that you can play with it

- anonymous

Alright, cool, thanks :)

- ganeshie8

Here it is
http://tube.geogebra.org/m/1437801

- ganeshie8

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

- anonymous

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.

- ganeshie8

if psble pls share, id like to see

- anonymous

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.

- ganeshie8

looks okay to me!

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