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#1 h Mar 30, 2022, 5:35 PM • 1 Y

Y by HWenslawski

Given a triangle $ABC$ with a circumcircle $\Gamma$
$AF$ is a symmedian line of triangle $ABC$
$A'$ is the symmetric point of $A$ wrt line $BC$
Point $D$ lies on the arc $AF$ of $\odot(AFA')$ (the one that does not contain $A'$ )
Point $D'$ is the intersection of AD and $\Gamma$
Prove that: $\angle DBD' =\angle DCD'$

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PROF65

#2 h Mar 30, 2022, 8:25 PM • 1 Y

Y by Wzy-Will

Wzy-Will wrote:

Given a triangle $ABC$ with a circumcircle $\Gamma$
$AF$ is a symmedian line of triangle $ABC$
$A'$ is the symmetric point of $A$ wrt line $BC$
Point $D$ lies on the arc $AF$ of $\odot(AFA')$ (the one that does not contain $A'$ )
Point $D'$ is the intersection of AD and $\Gamma$
Prove that: $\angle DBD' =\angle DCD'$

to prove the result it suffices to prove that the bisectors $DL,D'N$ of $\angle BDC ,\angle BD'C$ :
Let the $A$ -apollonius circle $(AA'F)$ cuts the segment $BC$ at $L$ ;
the $BC$ bisector cuts $(ABC)$ at $N,M$ where $N$ is the midpoint of the arc \overarc {BAC}
$d$ the tangent to $(AA'F)$ at $L$ which is perpendicular to $BC$ since the circle has its center on $BC$ ;
$\angle (d,LD)=\angle LAD=\angle MAD'=\angle MND'$ but $MN\parallel d$ hence $LD \parallel D'N$
My regards.
RH HAS

This post has been edited 2 times. Last edited by PROF65, Mar 31, 2022, 2:05 PM
Reason: add problem text

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#3 h Mar 31, 2022, 12:44 PM

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What's the definition of the A-aopllonius circle.

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PROF65

#4 h Mar 31, 2022, 1:20 PM

Y by

search it at Wolfram or Wikipedia.....

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#5 h Mar 31, 2022, 11:59 PM • 1 Y

Y by Wzy-Will

Reworded wrote:

Triangle $ABC$ has circumcircle $\Gamma$ . The $A$ -symmedian of $ABC$ meets $\Gamma$ again at $F$ . Let $A'$ be the reflection of $A$ over $BC$ . Point $D$ lies on arc $AF$ (not containing $A'$ ) of $(AFA'),$ and $D'$ is the second intersection between $AD$ and $\Gamma$ . Prove that $\angle DBD' =\angle DCD'$ .

Let $C_1$ be the reflection of $C$ over $AD$ , $T = AA \cap BC \cap FF$ , $K = AC \cap TD$ , and $X = AC_1 \cap TD$ . By symmetry, we have $TA = TA'$ , and equal tangents implies $TA = TF$ , so $T$ is the center of $(AFA')$ .

Because $AD$ bisects $\angle CAC_1 \equiv \angle KAX$ , $\angle TAX = \angle TAD - \angle XAD = \angle TDA - \angle KAD = \angle AKD = \angle AKX$ so $TA$ is tangent to $(AKX)$ . This yields $TK \cdot TX = |Pow_{(AKX)}(T)| = TA^2 = |Pow_{\Gamma}(T)| = TB \cdot TC$ so $BCKX$ is cyclic. Hence, we know $\angle BD'D = \angle BD'A = \angle BCA = \angle BCK = 180^{\circ} - \angle BXK = 180^{\circ} - \angle BXD$ implying $BXDD'$ is cyclic.

Now, observe $\angle XC_1D' = \angle AC_1D' = \angle ACD' = \angle TAD' = \angle TAD = \angle TDA$ $= \angle XDA = 180^{\circ} - \angle XDD'$ so $C_1XDD'$ is cyclic. Thus, $C_1BXDD'$ is a cyclic pentagon, which gives $\angle DBD' = \angle DC_1D' = \angle DCD'$ as desired. $\blacksquare$

Remark: It's well-known that $(AFA')$ is the $A$ -Apollonian Circle wrt $BC$ .

This post has been edited 1 time. Last edited by ike.chen, Apr 7, 2022, 4:44 AM
Reason: Absolute Value

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#6 h Apr 1, 2022, 3:41 AM • 1 Y

Y by ike.chen

Thanks very much indeed

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