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still need additional help on p18 in winter camp geometry blue

 
 
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still need additional help on p18 in winter camp geometry blue
by Iris Xu - Tuesday, 19 January 2021, 8:19 PM
 

I was not able to prove  triangle BGH is identical to triangle CGE, and why HGCOHGCO is a cyclic quadrilateral. also not sure how to prove CEF is similar to EBF. thanks


CircleTangentsSoln_9126.png

 
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Re: still need additional help on p18 in winter camp geometry blue
by Iris Xu - Sunday, 24 January 2021, 2:59 PM
 

Can a teacher please answer my question that was posted 5 days ago? thanks

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Re: still need additional help on p18 in winter camp geometry blue
by Dr. Kevin Wang - Tuesday, 26 January 2021, 7:24 PM
 

This question is one of the harder ones and it is hard to give hints without giving out the the full solution.  I'll give a solution in a few days.

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Re: still need additional help on p18 in winter camp geometry blue
by Iris Xu - Thursday, 28 January 2021, 1:45 AM
 

Thanks, professor. I look forward to your solution.

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Re: still need additional help on p18 in winter camp geometry blue
by Iris Xu - Friday, 29 January 2021, 5:16 PM
 

Hi professor, could you please post the solution now? thanks

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Re: still need additional help on p18 in winter camp geometry blue
by Dr. Kevin Wang - Sunday, 31 January 2021, 11:56 AM
 
This problem was one of the ZIML Varsity problems during 2018-2019.  Use the diagram above.


 Question: 

As shown in the diagram, $\overline{AB}$ and $\overline{AC}$ are two tangent lines of $\odot O$. Let $\overline{CD}$ be an angle bisector of $\triangle ABC$. Let $F$ be a point on the extension of $\overline{BC}$ such that $\overline{DF}$ is tangent to $\odot O$ at $E$. Given that $BC=21$, find $CF$. 


Solution: 

Let $G$ be the intersection of $\overline{CD}$ and $\odot O$, and connect $\overline{OB}$, $\overline{OC}$, $\overline{OG}$, $\overline{OD}$, $\overline{BG}$, $\overline{BE}$, and $\overline{CE}$. Let $H$ be the intersection of $\overline{BE}$ and $\overline{OD}$, and connect $\overline{HG}$, as shown in the diagram. 

 Since $\overline{BD}$ and $\overline{DE}$ are tangent to $\odot O$, $\overline{OD}\perp\overline{BE}$ and $H$ is the midpoint of $\overline{BE}$. Also $\angle OBD=90^\circ$, so $\triangle OBD\sim\triangle BHD$, and then $\dfrac{BD}{DH}=\dfrac{OD}{BD}$, which means $BD^2 = DH\cdot DO$. 

 Since $BD^2=DG\cdot DC$, we obtain \[ DH\cdot DO = DG\cdot DC, \] therefore $HGCO$ is a cyclic quadrilateral, and hence $\angle DHG=\angle DCO=\angle CGO$. Furthermore, we know that $\overline{CD}$ bisects $\angle ACB$, so $G$ is the midpoint of arc $\widehat{BC}$, then \[ \begin{split} \angle GHE=90^\circ-\angle DHG=90^\circ-DCO\\ =\angle DCA=\angle GCB=\angle GEH, \end{split} \] thus $GH=GE$. Also, \[ \angle HGE=180^\circ-2\angle GHE=180^\circ-2\angle GCB=\angle BGC, \] so \[ \angle BGH=\angle BGC-\angle HGC=\angle HGE-\angle HGC=\angle CGE. \] 

Now we have $GH=GE$, $\angle BGH=\angle CGE$, and $BG=CG$, so $\triangle BGH\cong\triangle CGE$ by $SAS$ congruency. Therefore $CE=BH=\dfrac{1}{2}BE$. Since $\angle CEF=\angle EBF$, we have $\triangle CEF\sim\triangle EBF$, and \[ \frac{CF}{EF} = \frac{EF}{BF} = \frac{CE}{BE} = \frac{1}{2}, \] thus \[ \frac{CF}{BF} = \frac{1}{4}, \] therefore \[ CF=\frac{1}{3}BC=7. \]

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Re: still need additional help on p18 in winter camp geometry blue
by Iris Xu - Monday, 1 February 2021, 10:41 PM
 

Thanks a lot, Professor!