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in a triangle ABC, P, Q are points on AB and R, S are points on BC such that AP=PQ=QB and CR=RS=SB. Show that PR bisects AS.

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    $\begingroup$ It is straightforward since PR is parallel to AC. Since S is not on AC, then AS is not parallel to AC and will interesect every parallel to AC... $\endgroup$ – Martigan Aug 11 '14 at 13:05
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    $\begingroup$ $PR$ is parallel to the base of the triangle $ACS$ at half its height. $\endgroup$ – Christian Blatter Aug 11 '14 at 14:21
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Here's a diagram ($I$ is the intersection of $\vec{PR}$ and $\vec{AS}$: enter image description here

Note that if a vector is a positive scalar multiple of another, they are parallel. We will prove that $\Delta ASC$ is similar to $\Delta ISR$. Obviously $\vec {SR}$ is parallel to $\vec {SC}$ and $\vec {IS}$ is parallel to $\vec {AS}$. We just have to prove that $\vec {IR}$ is parallel to $\vec {AC}$. This is as follows:

$\vec {AC}=\vec{AP}+\vec{PQ}+\vec{QB}+\vec{BS}+\vec{SR}+\vec{RC}=3(\vec{QB}+\vec{BS})$

$\vec {PR}=\vec{PQ}+\vec{QB}+\vec{BS}+\vec{SR}=2(\vec{QB}+\vec{BS})$

So $\vec {PR}$ is parallel to $\vec {AC}$. Since $\vec {PR}$ is obviously parallel to $\vec {IR}$, we now know that $\vec{IR}$ is parallel $\vec{AC}$. Since all three sides of $\Delta ISR$ are parallel to a side of $\Delta ASC$, those two triangles are similar. Therefore their ratio of sides is similar. We have

$\frac{|\vec{AS}|}{|\vec{IS}|}=\frac{|\vec{SC}|}{|\vec{SR}|}=\frac{2|\vec{SR}|}{|\vec{SR}|}=2\implies |\vec{AS}|=2\:|\vec{IS}|$.

That completes the proof.

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A proof with vectors... even if I find difficult to find the need of such a proof...

$\vec{AS} \otimes \vec{PR}=(\vec{AC}+2/3\vec{CB})\otimes 2/3 \vec{AC}$ (Thalès...)

Since $\vec{AC} \otimes \vec{AC}=\vec{0}$

$\vec{AS} \otimes \vec{PR}=4/9 \vec{CB}\otimes \vec{AC}\neq \vec{0}$ since $ABC$ is a regular triangle

$(AS)$ and $(PR)$ intersects

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