Let Γ be the circumcircle of ΔABC and let D be the midpoint of the arc BC. Prove that below three points are collinear :
1) A
2) the incenter I of ΔABC
3) D

http://tube.geogebra.org/m/1437801

Question

Let Γ be the circumcircle of ΔABC and let D be the midpoint of the arc BC. Prove that below three points are collinear :
1) A
2) the incenter I of ΔABC
3) D

http://tube.geogebra.org/m/1437801

ganeshie8 · Answer

Exactly, but somehow I find below statement not so obvious

"D lies on the angle bisector of angle A"

just looking for a proof if it is easy :)

loser66 · Answer

why not??

loser66 · Answer

$\angle BAD =\angle CAD$ since arc BD = arc CD

loser66 · Answer

hence AD bisects BAD

ganeshie8 · Answer

I can't answer "why not" because idk lol
By definition, $I$ lies on the angel bisector... but not so much about the point $D$ is known right

ganeshie8 · Answer

Ahh wait I see what you're doing, you're saying "equal chords of the same circle intercept equal angles" ?

loser66 · Answer

I don't know the name of the property but it is!! if the arcs of the angles are equal, then the angles are equal.

ganeshie8 · Answer

That should do! thanks!

loser66 · Answer

However, the original problem is not a piece of cake. I didn't find out the logic yet!! ha!! 
Can you please help me draw out the circle center D that goes through B? I need it to prove O is the incenter of $\triangle ABC$. Please

ganeshie8 · Answer

It is already there but hidden... just scroll down on the left hand side, do you see "Conic" section ?

ganeshie8 · Answer

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