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### still need additional help on p18 in winter camp geometry blue

still need additional help on p18 in winter camp geometry blue

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 Re: still need additional help on p18 in winter camp geometry blue

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

Re: still need additional help on p18 in winter camp geometry blue

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.

Re: still need additional help on p18 in winter camp geometry blue

Thanks, professor. I look forward to your solution.

Re: still need additional help on p18 in winter camp geometry blue

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

Re: still need additional help on p18 in winter camp geometry blue

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 $CF=\frac{1}{3}BC=7.$

Re: still need additional help on p18 in winter camp geometry blue

Thanks a lot, Professor!