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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Tags: geometry, circumcircle, parallelogram, cyclic quadrilateral, geometric transformation, geometry proposed
Luis González
23.01.2011 05:18
$(O)$ 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 $\angle PNA=\angle DCA=\angle AMB.$
FantasyLover
23.01.2011 07:41
We claim that $\triangle ABM\sim \triangle APN$. According to the property of spiral similarity, it is enough to prove that $\triangle ABP\sim \triangle AMN$. Since $\angle BAP=\angle MAN$, we will show that $ \angle ABP=\angle AMN$.
Choose point $X\in AM$ such that $MB^2=MC^2=MX\cdot MA$. Clearly, $\triangle MXC\sim \triangle MCA$, and this yields $\angle MCX=\angle MAC=\angle PCA$. Along with $\angle BAP=\angle CAX$, this implies that $P$ and $X$ are isogonal conjugates with respect to $\triangle ABC$.
Since $MN\parallel AB$ and $\triangle MXB\sim \triangle MBA$, $\angle AMN=\angle BAM=\angle XBM=\angle XBC$. On the other hand, since $P$ and $X$ are isogonal conjugates, $\angle XBC=\angle ABP$. Hence, $\angle ABP=\angle AMN$, and we are done. $\blacksquare$
dgreenb801
24.01.2011 05:10
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 $\angle PNA= \angle AMB$.
mathuz
24.04.2013 15:02
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.
sunken rock
25.04.2013 01:56
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 $\angle CPD=\angle CAP$. Best regards, sunken rock