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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|

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|

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

|dw:1437880661407:dw|

|dw:1437881020931:dw|

|dw:1437881129726:dw|

|dw:1437881183330:dw|

equaliteral triangle so we are gone i think

done

I is incenter dear, not circumcenter

|dw:1437881398303:dw|

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

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

Hmm didn't get you

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

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

Right :)

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 ?

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

if psble pls share, id like to see

looks okay to me!