If \angle BAO = \angle CAO, then \angle PAO = \angle QAO

kunny

#1

If \angle BAO = \angle CAO, then \angle PAO = \angle QAO

**Posted:** Sat Feb 21, 2009 6:37 pm

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cyshine · Poincare Conjecture Offline Joined: 08 Jun 2004 Posts: 176

#2

Two hints:

It is a known fact that...

Then prove that...

**Posted:** Sat Feb 21, 2009 8:25 pm

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campos

#3

let $M$ be the midpoint of arc $BC$ not containing $A$ , so $A,O,M$ are collinear...

let $O'$ be the center of $\Gamma$ , so $O', O, Q$ are collinear, and $OP|| O'M$ ...

then, $\angle QAO=\angle QAM=\frac{1}{2}\angle QO'M=\frac{1}{2}\angle POO'= \angle QPO$ ...

this implies that $APOQ$ is cyclic, and since $PO=OQ$ we conclude that $\angle PAO=\angle QAO$ ... Mr. Green

**Posted:** Mon Feb 23, 2009 5:37 am

livetolove212

#4

Change the supposition we have problem:
(O) is a circumcircle of $\Delta ABC$ and $AD,BM,CN$ are 3 bisectors. $(I),(J),(K)$ touch to line segment $BC,CA,AB$ at $P,E,F$ and touch the arc BC which doesn't have A, the arc CA which doesn't have B,the arc AB which doesn't have C at $Q,E,F$ . Prove that $AQ,BE,CF$ are concurrent Mr. Green

**Posted:** Sat Feb 28, 2009 7:37 am

livetolove212

#5

Hey who can solve my problem? Mr. Green

**Posted:** Sat Feb 28, 2009 3:22 pm

dgreenb801

#6

**Posted:** Sun Mar 15, 2009 11:47 pm

livetolove212

#7

**Posted:** Mon Mar 16, 2009 4:47 pm

quykhtn-qa1

#8

**Posted:** Tue Mar 17, 2009 7:10 am

cyshine · Poincare Conjecture Offline Joined: 08 Jun 2004 Posts: 176

#9

Hi dgreenb801,

Sorry I didn't answer your question before. I know livetolove212 have already proved it, but I have a different proof.

Consider the homothety with center $Q$ that takes the circle with center $O$ to the circumcircle. It also takes $BC$ to a line tangent to the circumcircle parallel to $BC$ , which touches it in the midpoint of the arc $BC$ (just do some angle chasing). This tangency point is the image of $P$ under such homothety, so it belongs to line $PQ$ .

**Posted:** Wed Mar 18, 2009 5:57 am

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