Triangle

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Marina19
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Triangle

Unread post by Marina19 » Sun Oct 07, 2012 3:21 pm

Hello!

I have a problem with this one:

Points P, Q, R lie respectively on sides BC, CA and AB of ABC triangle. AR=RP=PC and BR=RQ=QC. Prove that AC + BC =2 AB

Thanks for help in advance

Marina :)

Marina19
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Re: Triangle

Unread post by Marina19 » Tue Oct 16, 2012 10:00 pm

Maybe somehow laws of sines and cosines can be used here to prove it?

sakibtanvir
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Re: Triangle

Unread post by sakibtanvir » Tue Oct 16, 2012 11:06 pm

Is $\Delta ABC$ acute angled or it's not mentioned in the question? I asked this because the proof can be easily obtained then. :)
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Marina19
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Re: Triangle

Unread post by Marina19 » Tue Oct 16, 2012 11:47 pm

It's not mentioned. But can you share the proof for the acute triangle anyways? Maybe it can help in proving it in general.

sakibtanvir
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Re: Triangle

Unread post by sakibtanvir » Wed Oct 17, 2012 1:48 pm

I have found a generalized proof already. :D
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sakibtanvir
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Re: Triangle

Unread post by sakibtanvir » Wed Oct 17, 2012 2:16 pm

At first prove that,$PCQR$ is a rhombus. (Hint: Use congruence of triangles and the Angle Bisector Theorem)
$\Delta PQR \cong \Delta PCQ$.[Because,$RQ=QC,PR=PC$ and $PQ$ is a common side.].Let,$O$ be the intersection point of $RC$ and $PQ$.
$\therefore \angle RPQ=\angle CPQ,\angle RQP=\angle CQP$.In $\Delta PRC,PR=PC$ and $PO$ bisects $\angle RPC$.
So,$PO\perp RC,RO=OC$,Similarly taking the $\Delta RQC$,we can show that,$OQ\perp RC$.Which implies that,$PQ\perp RC$,implies that,$\Delta ROQ \cong \Delta COQ$.Now it is easy to prove
$\Delta ROP \cong \Delta COQ$,so $PR=QC$,Which is enough to prove it because $PR=PC$
Then,we are halfway there. :mrgreen: .
As, $PQ \parallel PC , RP \parallel QC$ ,We have,
$\frac{AR}{BR}=\frac{PC}{BP}=\frac{AQ}{QC}$.........(1)
But,$\frac{AR}{BR}=\frac{PR}{QC}=1$.................. (2)
Combining (1) & (2),$AR=BR,BP=PC,AQ=QC$.So, $P,Q,R$ are the midpoints of $BC,CA,AB$ respectively.
Again,$AR=PC \Rightarrow AB=BC$ and $QC=BR \Rightarrow AB=AC$.$\therefore AB=BC=CA \Rightarrow AC+BC=2AB$.
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Marina19
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Re: Triangle

Unread post by Marina19 » Wed Oct 17, 2012 8:00 pm

sakibtanvir wrote:Now it is easy to prove
$\Delta ROP \cong \Delta COQ$,so $PR=QC$,
Why is that>? I can't really see how to prove it

opilo
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Re: Triangle

Unread post by opilo » Wed Oct 31, 2012 11:51 pm

But it's exercise from this year Olimpiad Mathematic. Don't answer.

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Phlembac Adib Hasan
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Re: Triangle

Unread post by Phlembac Adib Hasan » Thu Nov 01, 2012 10:45 am

Marina19 is right. @Sakibtanvir, your proof is incorrect. From your argument, $PQRC$ doesn't need to be a rhombus.
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Nadim Ul Abrar
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Re: Triangle

Unread post by Nadim Ul Abrar » Thu Nov 01, 2012 3:45 pm

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Draw a parellal line $l$ to $QP$ through the point $R$ so that $l \cap AC=D$,$l\cap AB=E$.
its easy to prove that $Q,P$ are midpoints of $CD,CE$ respectively

Using sine rule in $\triangle ADR$ and $\triangle BER$


$\frac {AD}{BE}=\frac{\frac {AR}{sin \angle CDE}} {\frac {BR}{sin \angle CED}}$

$\frac{\frac {AR}{sin \angle CDE}} {\frac {BR}{sin \angle CED}}=1$ is directly followed from sine rule in $\triangle CED$

So $AD=BE$



Now $AC+BC=(QC+PC)+(QD+PB)=AB+(QD-AD+PE+BE)
=AB+(QD+PE)=(QC+PC)+AB=2AB$
$\frac{1}{0}$

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