Prove that ∠PNA = ∠AMB - [UKRMO 2009 Grade 9]

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Prove that ∠PNA = ∠AMB - [UKRMO 2009 Grade 9]

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#1 h Jan 23, 2011, 1:44 AM • 3 Y

Y by amatysten, Adventure10, Mango247

In triangle $ABC$ points $M, N$ are midpoints of $BC, CA$ respectively. Point $P$ is inside $ABC$ such that $\angle BAP = \angle PCA = \angle MAC .$ Prove that $\angle PNA = \angle AMB .$

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#2 h Jan 23, 2011, 3:18 AM • 2 Y

Y by Adventure10, Mango247

$(O)$ is circumcircle of $\triangle ABC$ and tangents through $A,B$ to $(O)$ cut the tangent through $C$ at $Q,R.$ A-symmedian $AR$ cuts $(O)$ again at $D$ and $P'$ is the midpoint of the chord $AD,$ i.e. projection of $O$ on $AD.$ Thus, $P'$ lies on the circumcircle of the cyclic quadrilateral $OCRB.$ Let $\odot(OBC)$ cut $AC$ again at $S.$ Then $\triangle ASB$ is isosceles with apex $S$ and $SR$ bisects $\angle BSA$ externally $\Longrightarrow$ $SR \parallel AB$ $\Longrightarrow$ $\angle BAP=\angle PRS=\angle PCA$ $\Longrightarrow$ $P \equiv P'.$ Now, since $NP$ is the A-midline of $\triangle ADC,$ it follows that $\angle PNA=\angle DCA=\angle AMB.$

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#3 h Jan 23, 2011, 5:41 AM • 1 Y

Y by Adventure10

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#4 h Jan 24, 2011, 3:10 AM • 4 Y

Y by zarengold, malilim, Adventure10, Mango247

Let $D$ be such that $ABDC$ is a parallelogram, then $A$ , $M$ , and $D$ are collinear. We have $\angle BAM= \angle PAC$ and $\angle BDA= \angle DAC= \angle ACP$ , so $\triangle ABD \sim \triangle APC$ . But $M$ is the midpoint of $AD$ and $N$ is the midpoint of $AC$ , so $\triangle ABM \sim \triangle APN$ . So $\angle PNA= \angle AMB$ .

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#5 h Apr 24, 2013, 12:02 PM • 2 Y

Y by Adventure10, Mango247

Let points $D, B$ lie opposite sides from the line $AC$ and $\triangle ABC$ similar to $\triangle DCA.$
That $ADCP$ is cyclic and $D, N, P$ are collinear. From
angles $DNC$ and $AMB$ are equal then $\angle ANP$ equal to $\angle AMB.$
Also solution the problem Grade 11, P:3 by the method is very easy.

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#6 h Apr 24, 2013, 10:56 PM • 1 Y

Y by Adventure10

Obviously, $P$ lies onto $A-$ symmedian of $\Delta ABC$ , and let $AP$ intersect the circle $\odot (ABC)$ at $D$ . Clearly, $ABDC$ is a harmonic quadrilateral and $\widehat{AMB}=\widehat{BMD}=\widehat{ACD}$ . Consequently we need to prove $\angle ANP=\angle ACD$ , i.e. $P$ is midpoint of $AD$ , but this is true, since $\angle CPD=\angle CAP$ .

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