At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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 :)
\(\angle BAD =\angle CAD\) since arc BD = arc CD
hence AD bisects BAD
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
Ahh wait I see what you're doing, you're saying "equal chords of the same circle intercept equal angles" ?
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.
That should do! thanks!
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
It is already there but hidden... just scroll down on the left hand side, do you see "Conic" section ?
got it!! thanks a lot
I got it!! lalala...
Ahhhh I missed the conversation because someone deleted their comments
Haha that's fine, I say irrelevant stuff all the time :P I kinda like distractions
Thats really a clever way to prove both the things one shot : I is the incenter and lies on the circle
I like the third paragraph in this: http://mathworld.wolfram.com/Collinear.html