Let ABC be a triangle with AB>AC. Let P be a point on line beyond A such that AP+PC=AB. Let M be the mid-point of BC and let Q be a point on the side AB such that CQ intersect AM at right angle. Prove that BQ = 2AP.

From the results in \(eq^{n}\) \([1 , 2 , 3]\), we get that \(DM = AP\) and \(\angle MDP = \angle APD\) => \(MDPA\) is isosceles trapezium. And hence, \(AM || PD\) ,also \(AM \perp PD\) => \(PD \perp QC\) but \(PD\) also bisects \(\angle APP'\)=> \(PQ = PC\) But also \(PB = PP'\). => \(BQ = CP' = 2AP\)

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Very nice solution sir .... :) Upvoted ✓

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Thanks. I appreciate you.

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@mithil shah It is not from RMO 2014. It is from INMO. I don't remember the Date.

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\(Constructions:::---\) Produce \(PC\) to \(P'\) such that \(CP'\) = \(2AP\). Join \(BP'\). Then draw \(PD \perp BP'\). Finally join \(MD\).

Given that- \(AP\) + \(PC\) = \(AB\)

=> \(2AP\) + \(PC\) = \(AB + AP\)

=> \(CP'\) + \(PC\) = \(PP'\) ---[Because \(CP'\) = \(2AP\).]

=> \(PB = PP'\)

=> \(\angle B = \angle P' =\theta\) And \(\angle BPD =\angle APD = 90-\theta\) ---- [\(1\)]

But \(PB = PP'\) and also \(PD \perp BP'\). => \(BD = DP'\)

=> \(DM\) = \(\frac{1}{2}CP'\) = \(AP\) ---- [\(2\)]& also, \(DM || P'C\) [By Midpoint Theorem]

\(DM || P'C\) => \(\angle PDM = 90 - \angle MDB\) = \(90 - \angle P' = 90 - \theta\). ---- [\(3\)]

From the results in \(eq^{n}\) \([1 , 2 , 3]\), we get that \(DM = AP\) and \(\angle MDP = \angle APD\) => \(MDPA\) is isosceles trapezium. And hence, \(AM || PD\) ,also \(AM \perp PD\) => \(PD \perp QC\) but \(PD\) also bisects \(\angle APP'\)=> \(PQ = PC\) But also \(PB = PP'\). => \(BQ = CP' = 2AP\)

\(K.I.P.K.I.G\)

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Maybe it is from INMO, i do not remember either.

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it is from CRMO-2 2014

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On which line is P; BA or CA?

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