DM = DN wanted, <CAP= < CBP, projections of interior P on BC, AC

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DM = DN wanted, <CAP= < CBP, projections of interior P on BC, AC

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Source: 2017 Moldova NMO 10.3

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#1 h Mar 17, 2021, 5:39 AM

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Let $P$ be a point, located inside a triangle $ABC$ , such that $\angle CAP= \angle CBP$ . Let $D$ be the midpoint of the side $AB$ , and $M$ and $N$ the projections of the point $P$ on the sides $BC$ and $AC$ , respectively. Prove that $DM = DN$ .

This post has been edited 1 time. Last edited by parmenides51, Mar 17, 2021, 5:47 AM

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#4 h Mar 17, 2021, 8:28 AM

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Cute.

Let $K=NP\cap (PMB)$ and $J=PM\cap (PNA)$ .
By Reim's theorem, we have $PN\perp AC\parallel BK$ and $PM\perp BC\parallel AJ$ .

Let $H=AC\cap BP$ and $G=BC\cap AP$ . By the angle condition $\angle CAP= \angle CBP$ , we have $AHGB$ is cyclic.

Therefore, we have $\measuredangle PAJ=\measuredangle PGB=\measuredangle AHP=\measuredangle KBP,$ hence $\triangle PAJ\sim\triangle PBK$ , therefore we have
$\measuredangle JNK=\measuredangle JAP=\measuredangle KBP=\measuredangle JMK,$ which implies that $JNMK$ is cyclic.
Now we look perpendicular bisectors of $JM$ and $NK$ , since $MJ\perp AJ\parallel BC$ , we have perpendicular bisector of $MJ$ bisecting $AB$ . Similarly, we conclude that perpendicular bisector of $NK$ bisects $AB$ , therefore the midpoint of $AB$ , $D$ , is the centre of $(JNMK)$ , we are done.

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#5 h Mar 17, 2021, 9:18 AM

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assume that $S$ and $T$ are midpoints of segments $PB$ and $PA$ respectively. so $DS=\frac{AP}{2}$ and $DT=\frac{PB}{2}$ . Also because

$\angle PMB=\angle PNA=90^{\circ}\Rightarrow$ $T$ and $S$ are circumcenter of triangles $NPA$ and $PMB$ . so $MS=\frac{PB}{2}$ and

$NT=\frac{AP}{2}$ . hence $MS=TD$ and $NT=DS$ . also by easy angle chasing and using the fact that $SD\parallel AP$ and $TD\parallel PB$ we

deduce that $\angle DSM =\angle DTN$ . hence $\triangle DSM \cong \triangle DTN \Rightarrow DM=DN$ and we are done!

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zuss77

#6 h Mar 17, 2021, 10:04 AM • 1 Y

Y by Mango247

This can't be original.
Let $Q$ be foot of $P$ on $AB$ . Isogonal conjugate of $P$ is on perp bisector of $AB$ , so $MNQD$ is cyclic (pedal circle). Also $QP$ bisects $\angle MQN$ (( $ANPQ$ ) and ( $BMPQ$ )). So $D$ - midpoint of the arc $MQN$ .

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