Prove that KL bisects BI

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Prove that KL bisects BI

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#1 h Mar 29, 2015, 4:34 AM • 1 Y

Y by Adventure10

Let $ABC$ is a triangle with circumcircle $(O)$ . Give a circle $(W)$ tangent $AB$ , $BC$ at $K$ , $L$ , resp, and tangent the circumcircle $(AOC)$ . Prove that $KL$ bisects $BI$ with $I$ is the centre of incircle $(ABC)$

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#2 h Mar 29, 2015, 5:30 AM • 2 Y

Y by Adventure10, Mango247

My solution:

Let $M \equiv BI \cap KL$ be the midpoint of $KL$ .
Let $A^*, C^*$ be the midpoint of $BC, BA$ , respectively .
Let $I_b$ be the B-excenter of $\triangle ABC$ and $O^*$ be the projection of $B$ on $AC$ .
Let $L^*, K^*$ be the projection of $I_b$ on $BA, BC$ , respectively .

Let $\Psi$ be the composition of Inversion $\mathbf{I}( \sqrt{\tfrac{1}{2} BA \cdot BC} )$ and Reflection $\mathbf{R} (BI)$ .

Since $A \longleftrightarrow A^*, C \longleftrightarrow C^*, O \longleftrightarrow O^*$ under $\Psi$ ,
so the image of $\odot (AOC)$ is the 9-point circle $\odot (A^*O^*C^*)$ of $\triangle ABC$ under $\Psi$ ,
hence the image of $\odot (W)$ is B-excircle $\odot (I_b)$ under $\Psi \Longrightarrow L\longleftrightarrow L^*, K \longleftrightarrow K^*$ under $\Psi$ ,
so we get $BL \cdot BL^*=BK \cdot BK^* =\tfrac{1}{2} BA \cdot BC$ .

Since $Rt \triangle BML \sim Rt \triangle BL^*I_b$ ,
so $BM \cdot BI_b=BL \cdot BL^*=\tfrac{1}{2} BA \cdot BC$ ,
hence combine with $BI \cdot BI_b= BA \cdot BC$ we get $M$ is the midpoint of $BI$ ,
so $KL$ is the perpendicular bisector of $BI$ .

Q.E.D

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andria

#3 h Mar 29, 2015, 8:34 AM • 2 Y

Y by Adventure10, Mango247

My solution: let the parallel line from $I$ to $KL$ meet $BA, BC$ at $S, T$ we want to prove that $KL$ is midline of $\triangle BTS$ note that $S, T$ are tangency points of mixtilinear circle in front of $A$ . apply an inversion with center $B$ and power $\sqrt { BC \cdot BA }$ we know that $C \longleftrightarrow A$ under the inversion and inverse of mixtilinear circle is excircle of $\triangle ABC$ in front of $\angle B$ that touches $BA, BC$ at $S', T'$ and since $ac=2Rh_B$ inverse of $O$ is a reflection of $A$ In the line $BC$ call it $O'$ let a line parallel to $BC$ from $O'$ intersect $BA, BC$ at $R, P$ so we get that $\odot ACO'$ is a nine point circle of $\triangle BRP$ also the inverse of $W$ is a circle that it is tangent to $BA, BC$ at $K', L'$ and tangent to $\odot \triangle AO'C$ ; call it $W'$ but note that from a well known fact the nine point circle of a triangle is tangent to its excircle from this fact we get that $W'$ is excircle of $\triangle BRP$ now because $A'C'$ is midline of $\triangle BRP$ we get that $T', S'$ are midpoints of $BL', BK'$ so $K, L$ are midpoints of $BS, BT$ before the inversion and $KL$ bisects $BI$ .

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#4 h Apr 9, 2015, 9:42 PM • 2 Y

Y by Adventure10, Mango247

See http://www.artofproblemsolving.com/community/c6h517026 for a solution without inversion.

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jayme

#5 h May 10, 2015, 11:56 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
a direct proof... based on the key circle (CMI)...

1. O' the midpoint of the arc BC which doesn’t contain O of (AOC)
M the second point of intersection de IO' with (O) ; it the contact point of (AOX) and (W)
(1) the circle (CMI)
X the point of intersection of KL and AI.
2. according to
http://jl.ayme.pagesperso-orange.fr/Docs/Un%20remarquable%20resultat%20de%20Vladimir%20Protassov.pdf p. 2-4
(1) goes through L and by angle chasing also through X
3. according to the Reim’s theorem, JL // AB
4. in the same way, JK // BC
5. the parallelogram BLIK is a rhombus and we are done.

Sincerely
Jean-Louis

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