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Refer to the figure below. The distances are $|AR| = 9, |RB| = 21, |BC| = 40, |CQ| = 18$, and $|QA| = 12$. Use Ceva’s Theorem for this configuration to calculate $|PC|$.

enter image description here

Curve's Theorem seems pretty straight forward, except I am not given $|BP|$. Here is what I have:

Attempt

\begin{align} \dfrac{AR}{RB}\cdot\dfrac{BP}{PC}\cdot\dfrac{CQ}{QA}=1&\Rightarrow \dfrac{BP}{PC}=\dfrac{252}{162}\\ &\Rightarrow\dfrac{BP}{PC}=\dfrac{14}{9}\\ &\Rightarrow PC=\dfrac{9BP}{14}\qquad(\star). \end{align} However, I cannot figure out how to find $BP$. My initial thought is to use the angle bisector theorem as follows:

\begin{align} \dfrac{BP}{PC}=\dfrac{BA}{CA}&\Rightarrow\dfrac{BP}{PC}=\dfrac{RA+RB}{AQ+QC}\\ &\Rightarrow\dfrac{BP}{PC}=\dfrac{30}{30}\\ &\Rightarrow\dfrac{BP}{PC}=1\\ &\Rightarrow BP=PC. \end{align} This now tells us that \begin{align} BP&=\dfrac{1}{2}BC\\ &=\dfrac{1}{2}40\\ &=20 \end{align} Substituting this result into $(\star)$ gives \begin{align} PC&=\dfrac{(9)(20)}{14}\\ &=12.86 \end{align} However, I am not sure this is correct. I am hoping someone can confirm correctness or point out where I have made errors.

Cheers

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  • $\begingroup$ You forgot about: $BP+PC=BC$ $\endgroup$
    – openspace
    May 10, 2016 at 1:54
  • $\begingroup$ And you have a mistake: you should make clockwise $\endgroup$
    – openspace
    May 10, 2016 at 2:06
  • $\begingroup$ If $BP=PC$ then $BP+PC=BC$ implies $2BP=BC$ which implies $BP=\dfrac{1}{2}BC$ is that not correct? Rather, my question is can I use the Bisector Theorem in this way? Also, I am unclear as to what you mean by "you should make clockwise". Can you clarify this please. Thanks $\endgroup$
    – Jeremy
    May 10, 2016 at 3:29
  • $\begingroup$ you may forgot about clockwise, actually. $BP+PC=BC$, but $PC=9BP/14$, so $BP+PC=23BP/14=BC$ $\endgroup$
    – openspace
    May 10, 2016 at 3:31
  • $\begingroup$ I mean : are you sure that this lines are bisectors ? $\endgroup$
    – openspace
    May 10, 2016 at 3:32

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