Midpoint of the internal bisector

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Midpoint of the internal bisector

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Source: ITAMO 2019 #5

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#1 h May 3, 2019, 4:30 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

Let $ABC$ be an acute angled triangle $.$ Let $D$ be the foot of the internal angle bisector of $\angle BAC$ and let $M$ be the midpoint of $AD.$ Let $X$ be a point on segment $BM$ such that $\angle MXA=\angle DAC.$ Prove that $AX$ is perpendicular to $XC.$

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#2 h May 3, 2019, 7:01 PM • 3 Y

Y by GGPiku, Pluto1708, Adventure10

Let $I$ be the incenter and $I_A$ be the excenter. The angle condition implies $MA$ is tangent to $(BXA)$ , so $MA^2 = MX \cdot MB$ . On the other hand, since $(A, D; I, I_A) = -1$ we have $MA^2 = MI \cdot MI_A$ . Hence $BXII_A$ is cyclic. But it is well-known that $C$ also lies on this circle. Now $\angle AXC = \angle A/2 + (180^{\circ} - \angle BXC) = \angle A /2 + (180^{\circ} - \angle BIC) = 90^{\circ}$ .

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jbaca

#4 h Aug 24, 2019, 1:40 AM • 2 Y

Y by A-Thought-Of-God, Adventure10

Solution. Let $I$ and $I_A$ be, as usual, the incenter and $A$ -excenter of $\bigtriangleup ABC$ . Since $(A,D;I,I_A)=-1$ and taking into account the angle condition, we get
$MX\cdot MB=MA^2=MI\cdot MI_A$ i.e. $IXBI_A$ is cyclic, thus
$\angle AXC=\angle MXA+\angle MXC=\angle BAD+180^\circ-\angle BIC=\angle BAD+90^\circ-\angle DAC=90^\circ$ as required. $\blacksquare$

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#5 h Aug 24, 2019, 2:07 AM • 2 Y

Y by A-Thought-Of-God, Adventure10

Essentially the same problem as 2014 Taiwan TST Round 1 P4.

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#6 h Oct 26, 2019, 9:09 PM • 2 Y

Y by Adventure10, Mango247

Let $AH\perp BC$ at $H$ . The condition implies $\angle MXD=\angle MDB$ . Then we can show that $X,M,D,H$ are concyclic by angle chasing. Thus $A,X,H,C$ are concyclic, and $AX\perp XC$ .

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Taha-R

#7 h Sep 26, 2023, 5:23 PM

Y by

If we draw a perpendicular from A to BC(point P), the result is that's enough to prove AXPC is cyclic.
$\Rightarrow$ It's enough to prove $\angle PXB = \angle ADB \Rightarrow$ It's enough to prove XMDP is cyclic $\Rightarrow$ It's enough to prove $\angle DXM = \angle DPM$ .
We khow that AM=MD=PM $\Rightarrow$ $\angle ADB = \angle DAC + \angle ACB = \angle DPM$ .
$\Rightarrow$ It's enough to prove $\angle DXM = \angle DAC + \angle ACB$ .
Claim1: $\angle ABM = \angle MAX$ .
Proof : $\angle AMB = \angle AMB$ and $\angle MXA = \angle MAB \Rightarrow MXA \sim MAB$ .
Claim2: $\angle MBD = \angle MDX$ .
Proof : By claim1 $AM^2 = MX.MB$ and AM=MD $\Rightarrow MD^2 = MX.MB \Rightarrow \angle MBD = \angle MDX$ .
$\Rightarrow \angle ABD = \angle XAM + \angle XDM$ , and the sum of the interior angles of the triangle AXD and ABD complete our proof.

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This post has been edited 2 times. Last edited by Taha-R, Sep 26, 2023, 5:27 PM
Reason: small problem !

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trying_to_solve_br

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trying_to_solve_br

#8 h Sep 26, 2023, 7:02 PM

Y by

Here's a nice sol.

Let $U$ be the meeting of $(ABX)$ with $AC$ and $Y=BX\cap AC$ . By POP and angle chase, $AU=AB$ and $\angle UXA=\angle AXY$ . Hence, as $XA$ bissects $UXY$ , we just need $(U,Y;A,C)=-1$ . But everything is now chaseable:

$AU/AY=CU/CY \iff AB/AB+AC=AY/CY$ . But $AY/CY$ , by sine law, is just $AB/BC . sin \angle ABY/sin \angle YBC$ .
But $sin \angle ABY/sin \angle YBC=sin ABM/sin MBD=BD/AB$ , again by sine law on $ABM,MBD$ using $MA=MD$ .
Thus, substituting everything we just need $AB/AB+AC=BD/BC \iff 1 + AC/AB=1+CD/DB=BC/BD$ , which is true by multiplying by $BD$ . Done =)

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#9 h Sep 27, 2023, 12:29 AM

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The condition implies $\triangle MAX \sim \triangle MBA$ , so $MX\cdot MB=MA^2=MD^2$ . Therefore $\overline{AMD}$ is tangent to $(BDX)$ . Let $E$ be the foot of the $A$ -altitude. Using $ME=MA=MD$ and the tangency, we have $\measuredangle MED=\measuredangle EDM=\measuredangle BXD=\measuredangle MXD$ , so $DEMX$ is cyclic. Then $\measuredangle CEX=\measuredangle DEX=\measuredangle DMX=\measuredangle MAX+\measuredangle AXM=\measuredangle CAM+\measuredangle MAX=\measuredangle CAX$ , hence $ACEX$ is cyclic so $\angle AXC=90^\circ.$ \blacksquare$

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