Triangle
Re: Triangle
Maybe somehow laws of sines and cosines can be used here to prove it?
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Re: Triangle
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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Re: Triangle
It's not mentioned. But can you share the proof for the acute triangle anyways? Maybe it can help in proving it in general.
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Re: Triangle
I have found a generalized proof already.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
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Re: Triangle
At first prove that,$PCQR$ is a rhombus. (Hint: Use congruence of triangles and the Angle Bisector Theorem)
Then,we are halfway there. .
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 $QC=BR \Rightarrow AB=AC$.$\therefore AB=BC=CA \Rightarrow AC+BC=2AB$.
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 $QC=BR \Rightarrow AB=AC$.$\therefore AB=BC=CA \Rightarrow AC+BC=2AB$.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
Re: Triangle
Why is that>? I can't really see how to prove itsakibtanvir wrote:Now it is easy to prove
$\Delta ROP \cong \Delta COQ$,so $PR=QC$,
Re: Triangle
But it's exercise from this year Olimpiad Mathematic. Don't answer.
- Phlembac Adib Hasan
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Re: Triangle
Marina19 is right. @Sakibtanvir, your proof is incorrect. From your argument, $PQRC$ doesn't need to be a rhombus.
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