Show that $OX=OY$. [closed]

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.

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

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

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
• Very educational answer ! – Jean Marie Nov 7 '18 at 14:53
• @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. – rafa Nov 9 '18 at 6:08
• @rafa since $AB$ is a tangent line, $\angle BAY = 1/2 \arc{ACY} = \pi - \angle ACY$. So $\angle XAB = \angle ACY$. Similarly, $\angle BXA = \angle CAY$. – Quang Hoang Nov 9 '18 at 6:13

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:

$$DX=DA=OE\tag{1}$$

$$DO=AE=EY\tag{2}$$

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)

Finally:

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

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