## 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.

1. ganeshie8

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

2. anonymous

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

3. 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..

4. anonymous

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

5. ganeshie8

Basically we want to show $$DB=DI=DC$$ |dw:1437879935160:dw|

6. anonymous

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

7. ganeshie8

Interesting, do we know if AID lie on a straight line from some theorem or somehting ? |dw:1437880349134:dw|

8. 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.

9. ganeshie8

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

10. ganeshie8

|dw:1437880661407:dw|

11. 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.

12. ganeshie8

@dan815

13. dan815

|dw:1437881020931:dw|

14. dan815

|dw:1437881129726:dw|

15. dan815

|dw:1437881183330:dw|

16. dan815

equaliteral triangle so we are gone i think

17. dan815

done

18. dan815

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

19. ganeshie8

I is incenter dear, not circumcenter

20. 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.

21. dan815

|dw:1437881398303:dw|

22. 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.

23. ganeshie8

Gotcha!

24. ganeshie8

cyclic quadrilateral does the job !

25. anonymous

I was considering that also, but is it cyclic?

26. ganeshie8

|dw:1437881873780:dw|

27. anonymous

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

28. ganeshie8

|dw:1437881954866:dw|

29. anonymous

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

30. ganeshie8

bottom is an arbitrarily inscribed angle top is fixed, incenter

31. dan815

ok got it

32. 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

33. 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

34. anonymous

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

35. ganeshie8

Hmm didn't get you

36. 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.

37. ganeshie8

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

38. 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.

39. 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|

40. ganeshie8

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

41. anonymous

Right :)

42. 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

43. ganeshie8

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

44. anonymous

Will do.

45. ganeshie8

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

46. anonymous

Whoa, how'd you make that?

47. ganeshie8

familiar with geogebra ?

48. 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

49. ganeshie8

its pretty good and very easy to learn

50. ganeshie8

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

51. anonymous

Alright, cool, thanks :)

52. ganeshie8

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

53. 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

54. 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.

55. ganeshie8

if psble pls share, id like to see

56. 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.

57. ganeshie8

looks okay to me!