CD bisects arc EF

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CD bisects arc EF

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Source: ItaMO 2005, P6

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Sayan

#1 h Mar 9, 2012, 3:55 PM • 2 Y

Y by Adventure10, Mango247

Two circles $\gamma_1, \gamma_2$ in a plane, with centers $A$ and $B$ respectively, intersect at $C$ and $D$ . Suppose that the circumcircle of $ABC$ intersects $\gamma_1$ in $E$ and $\gamma_2$ in $F$ , where the arc $EF$ not containing $C$ lies outside $\gamma_1$ and $\gamma_2$ . Prove that this arc $EF$ is bisected by the line $CD$ .

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WakeUp

#2 h Mar 9, 2012, 5:10 PM • 2 Y

Y by Adventure10, Mango247

We'll prove that $D$ is the incentre of $\triangle CEF$ . Note that $\angle DAB=\angle CAB=\angle CFB=\angle FCB=\angle FAB$ so $A,D,F$ are collinear (since $D,F,$ lie on the same side of the line $BA$ ). By a symmetric argument, $B,D,E$ are collinear.

From here, I had originally applied inversion and followed through that way.. but it's actually quicker not to do so. Anyway $\angle CED=\angle CEB=\angle CFB=\angle BCF=\angle BEF=\angle DEF$ . So $DE$ is an angle bisector of $\angle CEF$ . It's also, by a symmetric argument, the angle bisector of $\angle CFE$ , so $D$ must be the incentre of $\triangle CEF$ . Hence $CD$ bisects the arc $EF$ not containing $C$ on $(CEF)$ .

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#3 h Aug 11, 2012, 6:36 AM • 2 Y

Y by Adventure10, Mango247

$\angle{ABC}=\angle{AEC}$
$\angle {ACG}=9O^0-\angle{CAB}=\angle{ECG}+\angle{ACB}$
So $\angle {ECG}=\angle {GCF}=x$
So $EG=2r Sin x=GF$
Done.

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#4 h Aug 11, 2012, 7:24 PM • 1 Y

Y by Adventure10

Let $\{F'\}\in AD\cap\gamma_2, \{E'\}\in BD\cap\gamma_1$ . Triangles $\triangle BDF',\triangle ADE'$ are isosceles and $\angle DF'B=\angle BDF'=180^\circ-\widehat{ADB}$ , but $\angle ADB=\angle ACB$ , so $BCAF'$ is cyclic, so $F'\equiv F$ ; similarly $E'\equiv E$ . Since $AC=AE\implies AF$ is bisector of $\angle CFE$ , in a similar way $BE$ is bisector of $\angle FEC$ , consequently $D$ is the incenter of $\triangle FEC$ , done.

Best regards,
sunken rock

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#5 h Aug 16, 2012, 11:31 AM • 1 Y

Y by Adventure10

My solution:

We prove that $A$ , $D$ , $F$ are collinear and that $B$ , $D$ , $E$ are collinear as was done earlier in the thread. Then $\angle ECD = \frac{1}{2} \angle EAD = \frac{1}{2} \angle DBF = \angle DCF$ so $CD$ bisects arc $EF$ as required.

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