Not only X on A-altitude, but also AO_aA_aX parallelogram

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Not only X on A-altitude, but also AO_aA_aX parallelogram

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Source: 4th QEDMO problem 9, originally from John Sturgeon Mackay in Edinburgh Proceedings 1883

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#1 h Nov 10, 2007, 1:51 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle.
The $A$ -excircle of triangle $ABC$ has center $O_{a}$ and touches the side $BC$ at the point $A_{a}$ .
The $B$ -excircle of triangle $ABC$ touches its sidelines $AB$ and $BC$ at the points $C_{b}$ and $A_{b}$ .
The $C$ -excircle of triangle $ABC$ touches its sidelines $BC$ and $CA$ at the points $A_{c}$ and $B_{c}$ .
The lines $C_{b}A_{b}$ and $A_{c}B_{c}$ intersect each other at some point $X$ .
Prove that the quadrilateral $AO_{a}A_{a}X$ is a parallelogram.

Remark. The $A$ -excircle of a triangle $ABC$ is defined as the circle which touches the segment $BC$ and the extensions of the segments $CA$ and $AB$ beyound the points $C$ and $B$ , respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $CAB$ and the exterior angle bisectors of the angles $ABC$ and $BCA$ .
Similarly, the $B$ -excircle and the $C$ -excircle of triangle $ABC$ are defined.

Source of the problem

This post has been edited 1 time. Last edited by darij grinberg, Nov 11, 2007, 12:38 PM

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yetti

#2 h Nov 11, 2007, 1:50 AM • 2 Y

Y by Adventure10, Mango247

$O_aB \parallel A_bC_b,$ both perpendicular to the bisector $BO_b$ of the $\angle B,$ and $O_aC \parallel A_cB_c,$ both perpendicular to the bisector $CO_c$ of the $\angle C.$ The $\triangle O_aBC \sim \triangle A_bA_cX$ are centrally similar, having parallel sides. (Since $BA_c = s - a = CA_b$ and $A_bA_c = b + c,$ the similarity center is the common midpoint $A'$ of their corresponding sides $BC \sim A_bA_c$ and the similarity coefficient $\frac {a}{b + c}.$ ) Let $D \in BC$ be foot of the A-altitude and let $K \in BC$ be foot of the $\angle A$ bisector $AO_a$ of the $\triangle ABC.$ A is the internal similarity center of the excircles $(O_b),\ (O_c),$ $\frac {AO_b}{AO_c} = \frac {r_b}{r_c} = \frac {DA_b}{DA_c}.$ The points $A_a,\ K$ divide the side $BC$ of the $\triangle O_aBC$ in the ratios

$\frac {A_aB}{A_aC} = \frac {s - c}{s - b},\ \ \frac {KB}{KC} = \frac {c}{b},$

while the points $D,\ A_a$ divide the corresponding side $A_bA_c$ of the $\triangle XA_bA_c$ in the ratios

$\frac {DA_b}{DA_c} = \frac {r_b}{r_c} = \frac {s - c}{s - b},\ \ \frac {A_aA_b}{A_aA_c} = \frac {s - a + s - b}{s - a + s - c} = \frac {c}{b}.$

Thus $A_a \sim D,\ K \sim A_a$ are corresponding points of the centrally similar $\triangle O_aBC \sim \triangle A_bA_cX,$ therefore $O_aA_a \parallel XD,\ O_aK \parallel XA_a.$ $A \in O_aK$ by definition, $O_aA \equiv O_aK.$ $A \in XD,$ $XA \equiv XD,$ because $AD \perp BC$ by definition and $O_aA_a \perp BC,$ hence $XD \perp A_bA_c \equiv BC.$

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#3 h Nov 25, 2007, 10:48 AM • 2 Y

Y by Adventure10, Mango247

yetti wrote:

$O_aB \parallel A_bC_b,$ both perpendicular to the bisector $BO_b$ of the $\angle B,$ and $O_aC \parallel A_cB_c,$ both perpendicular to the bisector $CO_c$ of the $\angle C.$

Another solution of using this.
Denot $P,Q=l\cap XA_{b},XA_{c}$ where $l$ is line that pass $A$ and parallel to $BC$ . Then triangle $O_{a}BC$ and $XPQ$ are homothety. $BA_{b}=BC_{b}$ imply $AP=AC_{b}=A_{a}B$ , and similarly $AQ=A_{a}C$ . Hence $PQ=BC$ . Therefore, triangle $O_{a}BC$ and $XPQ$ are congruent and homothety.Since $AP=A_{a}B$ and $AQ=A_{a}C, A_{a},A$ are corresponding points.So $XA=O_{a}A_{a}$ and $XA\parallel O_{a}A_{a}$ .Hence $AO_{a}A_{a}X$ is parallelogram.

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