Let $O$ the circumcenter of an acute triangle $ABC$. Let $\alpha $ the circle trough $A $ and $B $ tangent to $[AC] $, and $\beta $ the circle trough $A, C $ tangent to $[AB] $. A line trough $A $ intersects the second time the circles $\alpha $ and $\beta $ in $X $ and $Y $. Show that $OX=OY $.

I try to check that the second point of intersection of the circles is $O $. With GeoGebra this fact is false.

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deleted by user21820, José Carlos Santos, Saad Mar 31 at 1:01

closed as off-topic by user21820, YiFan, Shailesh, Thomas Shelby, Parcly Taxel Mar 17 at 6:23

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  • $\begingroup$ If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing. $\endgroup$ – Quang Hoang Nov 7 '18 at 14:25

enter image description here

Since I’m on the phone and can’t type properly I will make some hints:

  1. Show that triangles $ABX$ and $ACY$ are similar.
  2. Show that $AEOD$ is a parallelogram.
  3. Show that triangles $XEO$ and $ODY$ are congruent.
deleted Mar 31 at 1:01
  • $\begingroup$ Very educational answer ! $\endgroup$ – Jean Marie Nov 7 '18 at 14:53
  • $\begingroup$ @QuangHoang Can you explain why $\triangle ABX$ ~ $\triangle CBY $? I can deduce from the second assertion which is very clear that $\angle AXB=\angle CBY $. I need to prove another congruence. $\endgroup$ – rafa Nov 9 '18 at 6:08
  • $\begingroup$ @rafa since $AB$ is a tangent line, $\angle BAY = 1/2 \arc{ACY} = \pi - \angle ACY$. So $\angle XAB = \angle ACY$. Similarly, $\angle BXA = \angle CAY$. $\endgroup$ – Quang Hoang Nov 9 '18 at 6:13

enter image description here

My proof does not use similarity. So it's not just a copy of Quang's (excellent) idea.

Notice that:

$$OD\bot AB,\space EA\bot AB \implies DO\parallel AE$$

$$OE\bot AC,\space DA\bot AC \implies DA\parallel OE$$

Obviously, quadrilateral ADOE is a parallelogram (opposite sides are parallel) and because of that:



Introduce angles $\delta=\angle EAY$, $\gamma=\angle BAC$

$$\angle AEY=180^\circ-2\delta$$

$$\angle AEO=\gamma$$

(the last is true because angles ∠AEO and ∠BAC have perpendicular legs)

$$\angle OEY=360^\circ-\angle AEY -\angle AEO=180^\circ+2\delta-\gamma\tag{3}$$

On the other side opposite angles in parallelogram $\angle DAE$ and $\angle DOE$ are equal and $\angle DOE$ is obtuse angle with perpendicular legs with respect to angle $\angle BAC=\gamma$:

$$\angle DAE=\angle DOE=180^\circ-\gamma$$

$$\angle DAX=180^\circ-\angle DAE-\angle EAY=\gamma-\delta$$

$$\angle DXA=\angle DAX=\gamma-\delta$$

$$\angle XDA=180^\circ-\angle DAX-\angle DXA=180^\circ-2\gamma+2\delta$$

$$\angle ADO=\gamma$$

(the last is true because angles $\angle ADO$ and $\angle BAC$ have perpendicular legs)


$$\angle XDO=\angle XDA+\angle ADO=180^\circ+2\delta-\gamma\tag{4}$$

From (3) and (4):

$$\angle XDO=\angle OEY\tag{5}$$

Because of (1), (2) and (5) triangles $\triangle XDO$ and $\triangle OEY$ are congruent by SAS so it must be that $OX=OY$.

deleted Mar 31 at 1:01

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