The line MN bisects the segment \overline{AH}

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sqing

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sqing

#1 h Dec 25, 2018, 10:34 AM • 1 Y

Y by Adventure10

Let $ABC$ be an acute triangle with $|AB | < |AC |$ and orthocenter $H$ . The circle with center A and radius $|AC |$ intersects the circumcircle of $\triangle ABC$ at point $D$ and the circle with center $A$ and radius $|AB |$ intersects the segment $\overline{AD}$ at point $K.$ The line through $K$ parallel to $CD$ intersects $BC$ at the point $L.$ If $M$ is the midpoint of $\overline{BC}$ and N is the foot of the perpendicular from $H$ to $AL,$ prove that the line $MN$ bisects the segment $\overline{AH}$ .

This post has been edited 1 time. Last edited by sqing, Dec 25, 2018, 10:35 AM

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TheDarkPrince

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TheDarkPrince

#2 h Dec 25, 2018, 11:21 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

sqing wrote:

Let $ABC$ be an acute triangle with $|AB | < |AC |$ and orthocenter $H$ . The circle with center A and radius $|AC |$ intersects the circumcircle of $\triangle ABC$ at point $D$ and the circle with center $A$ and radius $|AB |$ intersects the segment $\overline{AD}$ at point $K.$ The line through $K$ parallel to $CD$ intersects $BC$ at the point $L.$ If $M$ is the midpoint of $\overline{BC}$ and N is the foot of the perpendicular from $H$ to $AL,$ prove that the line $MN$ bisects the segment $\overline{AH}$ .

Solution: As $\angle BLK = \angle BCD = \angle B - \angle C = \angle BAK$ which gives $A,B,K,L$ are concyclic and $AL$ is the angle bisector of $\angle BAC$ . Suppose $X$ is the midpoint of $AH$ and $O$ is the circumcenter of $\triangle ABC$ , then $\angle HXN = 2\angle HAN = \angle B- \angle C = \angle HAO$ which gives $XN||AO$ . We know that $AXMO$ is a parallelogram, therefore $X,M,N$ are collinear and we are done. $\square$

This post has been edited 2 times. Last edited by TheDarkPrince, Dec 25, 2018, 11:29 AM
Reason: added reason for angle bisector

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PROF65

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PROF65

#3 h Jan 18, 2019, 7:59 AM • 2 Y

Y by Adventure10, Mango247

Let $G$ be the midpoint of $AH$ . since $ANH$ is rectangle then $\angle HGN =2\angle HAN$ so to prove that $G,N,M$ are colinear it sufffices to prove that $AN$ is the bisector :
Let $S=(A,AB)\cap AC$ then $S \in KL$ then $\angle ASL =\angle ABC$ besides $AS=AB$ we conclude that $S$ is the symmetric of $B$ about $AL$ which means that $AL$ is the bisector of $\angle BAC$ .
RH HAS

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MathInfinite

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MathInfinite

#5 h Mar 31, 2019, 11:13 AM • 1 Y

Y by Adventure10

Also, see https://artofproblemsolving.com/community/u404655h1766909p11573999

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Ditzymathstar

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Ditzymathstar

#6 h Mar 31, 2019, 8:43 PM • 2 Y

Y by Adventure10, Mango247

once you get the fact that AL is the angle bisector, you can literally kill the problem by using barycentric..
You'll get all the coordinates easily considering ABC as the reference triangle.

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e_plus_pi

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e_plus_pi

#7 h Jul 27, 2019, 2:12 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

sqing wrote:

Let $ABC$ be an acute triangle with $|AB | < |AC |$ and orthocenter $H$ . The circle with center A and radius $|AC |$ intersects the circumcircle of $\triangle ABC$ at point $D$ and the circle with center $A$ and radius $|AB |$ intersects the segment $\overline{AD}$ at point $K.$ The line through $K$ parallel to $CD$ intersects $BC$ at the point $L.$ If $M$ is the midpoint of $\overline{BC}$ and N is the foot of the perpendicular from $H$ to $AL,$ prove that the line $MN$ bisects the segment $\overline{AH}$ .

Let $T$ denote the mid-point of $AH$ . We present the following claims:
$\text{[math]}$
Claim 1: $ALKB$ is a cyclic quadrilateral.
Proof. We proceed with elementary angle chasing: $\angle KAC = \angle DAC = 180^{\circ} - 2\angle B \implies \angle BAK = \angle BAD = \angle B - \angle C$ Also, $\angle BLK = \angle (BC , KL) = 180^{\circ} - [ 180^{\circ} - \angle B + \angle C] = \angle B - \angle C$ .
$\text{[math]}$
Claim 2 : $AL$ is the $A-$ internal angle bisector.
Proof. We have $\angle BAL = 180^{\circ} - \angle BKL = 180^{\circ}- [ 90^{\circ} - \frac{1}{2} \cdot (\angle B - \angle C) + \angle B] = 90^{\circ} - \frac{1}{2} \cdot (\angle B + \angle C) = \frac{1}{2} \cdot \angle A$
$\text{[math]}$
Claim 3: $T - N - M$
Proof. Observe that $\angle HTN = 2\cdot \angle HAL = \angle B - \angle C = \angle HAO \implies TN \parallel AO$ . It is also well-known that $AOMT$ is a $\parallel^{\text{gm}} \implies TM \parallel AO \implies T - N - M$ .

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AlastorMoody

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AlastorMoody

#8 h Sep 4, 2019, 1:57 PM • 1 Y

Y by Adventure10

EMC 2018 Juniors P3 wrote:

Let $ABC$ be an acute triangle with $|AB | < |AC |$ and orthocenter $H$ . The circle with center A and radius $|AC |$ intersects the circumcircle of $\triangle ABC$ at point $D$ and the circle with center $A$ and radius $|AB |$ intersects the segment $\overline{AD}$ at point $K.$ The line through $K$ parallel to $CD$ intersects $BC$ at the point $L.$ If $M$ is the midpoint of $\overline{BC}$ and N is the foot of the perpendicular from $H$ to $AL,$ prove that the line $MN$ bisects the segment $\overline{AH}$ .

Solution: WLOG, Assume $K$ inside $\Delta ABC$ . Since, $\angle BAK=\angle BCD=\angle KLC$ $\implies$ $ABLK$ is cyclic $\implies$ $AL$ bisects $\angle BLK$ $\implies$ $L$ lies on $A-$ angle bisector. Let $P$ be midpoint of $AH$ $\implies$ $P$ is center of $\odot (AHN)$ . Let $PM$ $\cap$ $AL$ $=$ $N'$ $\implies$ $\angle LN'M$ $=$ $\angle LAO$ $=$ $\angle PN'A$ $=$ $\angle HAL$ $\implies$ $\Delta PAN'$ is isosceles $\implies$ $N \equiv N'$ $\qquad \blacksquare$

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#9 h Dec 25, 2021, 5:50 PM

Y by

∠BAK = ∠BCD = ∠BLK ---> ABKL is cyclic.
Now let's prove AL is angle bisector of ∠BAC.
∠CAL = 180 - ∠ALB = 180 - ∠AKB = ∠BKD and ∠ACL = ∠KDB so ∠CAL = ∠DBK.
∠DBK = (180 - ACD) - ABK = (180 - (90 - ∠CAD/2)) - (90 - ∠DAB/2) = ∠CAD/2 + ∠DAB/2 = ∠BAC/2 so ∠CAL = ∠BAC/2.

Let S be midpoint of AH and O the circumcenter of ABC.
∠HAN = 2∠HAN = ∠B - ∠C = ∠HAO ---> NS || OA.
we know AH = 2OM so AS = MH so ASMH is parallelogram and MS || OA.
so M,N,S are collinear and we're Done.

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