Those lines are the symmedians of the triangle, so you bet they're concurrent! Try to show that $A_2$ is the intersection of the tangents in $B,C$ to the circumcircle $(O)$ of $ABC$.

paul_mathematics wrote: Given a nonisosceles, nonright triangle ABC, let O denote the center of its circumscribed circle, and let $A_1$, $B_1$, and $C_1$ be the midpoints of sides BC, CA, and AB, respectively. Point $A_2$ is located on the ray $OA_1$ so that $OAA_1$ is similar to $OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1$, respectively, are defined similarly. Prove that lines $AA_2$, $BB_2$, and $CC_2$ are concurrent, i.e. these three lines intersect at a point. Since the triangle $OAA_1$ is similar to triangle $OA_2A$, we have $\frac{OA_1}{OA} = \frac{OA}{OA_2}$. On the other hand, we have OA = OB, since the point O is the circumcenter of triangle ABC, and thus this rewrites as $\frac{OA_1}{OB} = \frac{OB}{OA_2}$. Together with $\measuredangle BOA_1 = \measuredangle A_2OB$, this yields that the triangles $OBA_1$ and $OA_2B$ are similar, what, in turn, shows us that $\measuredangle OBA_2 = \measuredangle OA_1B$. But the point O is the circumcenter of triangle ABC and thus lies on the perpendicular bisector of its side BC; since the point $A_1$ is the midpoint of this side BC, we thus have $OA_1\perp BC$ and $\measuredangle OA_1B = 90^{\circ}$. Consequently, $\measuredangle OBA_2 = \measuredangle OA_1B = 90^{\circ}$, and $BA_2 \perp OB$. Since OB is a radius of the circumcircle of triangle ABC, the line $BA_2$ is therefore the tangent to the circumcircle of triangle ABC at the point B. In other words, the point $A_2$ lies on the tangent to the circumcircle of triangle ABC at the point B. Similarly, the same point $A_2$ lies on the tangent to the circumcircle of triangle ABC at the point C. Thus, our point $A_2$ is the point of intersection of the tangents to the circumcircle of triangle ABC at the points B and C. Similarly, the point $B_2$ is the point of intersection of the tangents to the circumcircle of triangle ABC at the points C and A, and the point $C_2$ is the point of intersection of the tangents to the circumcircle of triangle ABC at the points A and B. Now, the concurrence of the lines $AA_2$, $BB_2$, $CC_2$ becomes a well-known fact (in fact, these lines concur at the symmedian point of triangle ABC); the simplest proof of this fact uses the Ceva theorem: Since the two tangents from a point to a circle are equal in length, we have $CB_2 = B_2A$, $AC_2 = C_2B$ and $BA_2 = A_2C$, so that $\frac{B_2A}{AC_2}\cdot\frac{C_2B}{BA_2}\cdot\frac{A_2C}{CB_2} = \frac{B_2A}{C_2B}\cdot\frac{C_2B}{A_2C}\cdot\frac{A_2C}{B_2A} = 1$. After the Ceva theorem, applied to the triangle $A_2B_2C_2$, it now follows that the lines $A_2A$, $B_2B$, $C_2C$ are concurrent. In other words, the lines $AA_2$, $BB_2$, $CC_2$ are concurrent. $\blacksquare$ As usual, Grobber was quicker than me . Anyway, I have tried to give a proof as elementary and detailed as possible since I don't know how much advanced geometry you know. I also have solutions for some of the other problems you posted here, but I am quite under time pressure now and I'll look whether I will succeed to write them up. Darij

Alternatively, you can draw the altitude $AH$. $AH \parallel OA_2$, so it's obvious that $\angle BAA_2 = \angle A_1AC$, so $AA_2$ is a symmedian, and similarly for $BB_2$, $CC_2$.

Let $G$ be the centroid and $H$ the orthocenter of $\triangle ABC$. Then $\angle OAA_2=\angle OA_1A=\angle A_1AH$, and $\angle BAO=90^{\circ}-C=\angle HAC$, so we have $\angle BAA_2=\angle A_1AC$. Similarly we can get $\angle AA_2C=\angle BAA_2$ and so on. By trig Ceva we have \begin{align*}\frac{\sin\angle BAA_2}{\sin\angle A_2AC}\cdot\frac{\sin\angle ACC_2}{\sin\angle C_2CB}\cdot\frac{\sin\angle CBB_2}{\sin\angle B_2BA}&=\frac{\sin\angle A_1AC}{\sin\angle BAA_1}\cdot\frac{\sin\angle B_1BA}{\sin\angle CBB_1}\cdot\frac{\sin\angle C_1CB}{\sin\angle ACC_1}\\ &=1\end{align*} since $AA_1$, $BB_1$, and $CC_1$ concur at $G$. Therefore $AA_2$, $BB_2$, and $CC_2$ are concurrent as well. $\Box$

Note that $(OA_1)(OA_2)=OA^2$, so $A_2$ is the image of $A_1$ about an inversion with respect to the circle. Because $BA_1C$ are collinear, $OBA_2C$ are concyclic ($B$ and $C$ are their own images under the inversion centered at $O$). Clearly there is a unique point $X$ that is not $O$ on $OA_1$ such that $OBXC$ are conyclic. Now, the intersection of the tangents to the circumcircle at $B$ and $C$ is clearly on $OA_1$ and clearly is concyclic with $COB$ since if that points is $X$, $<XCO=<XBO=90$. Hence, $A_2$ is the intersection of the tangents to the circumcircle at $B$ and $C$. Thus, $AA_2$ is a symmedian by the well known symmedian lemma. The others are similarly symmedians, so we are done and they concur by the isogonal conjugates theorem. As a consequence, we have that the inversions of the lines $AA_2$ which are the circles through $AOA_1$, $BOB_1$, $COC_1$, concur. In fact we have that they concur at the image of the symmedian point after an inversion about $O$.

clearly, the best way to do it is We use a polar transformation with respect to the circumscribed circle, so we need to prove that $BB \cap AC$, $CC \cap AB$, and $AA \cap BC$ are concurrent. But this is easy through Pascal's theorem on $AABBCC$

We'll angle chase to show that $AA_2,BB_2,CC_2$ are the symmedians of $\triangle ABC$. By a well known fact we know that those intersect in the Lemoine Point/Symmedian Point. (We could also easily prove it using Ceva by using the property that if $D$ lies on $BC$, such that $AD$ is the $A$-symmedian of $\triangle ABC$, then $\tfrac{|BD|}{|DC|} = \left( \tfrac{|AB|}{|CA|} \right)^2$.) Now let $\theta := \angle C_1CO$ and $\sigma := \angle OC_1C$. By the similarity we get $\angle C_2CO= \sigma$ and $\angle OC_2C = \theta$. By simple angle sum in the triangle we get \[ \angle C_1CB = 180^{\circ} - \beta - \left(90^{\circ}+\sigma \right) = 90^{\circ}-\beta-\sigma. \]We also get \[ \angle ACC_2 = \gamma - (\sigma-\theta)-\left(90^{\circ}-\beta-\sigma \right) = \beta+\gamma+\theta-90^{\circ} \]Thus, by the definition of the symmedian it suffices to prove \[ \angle ACC_2 = \angle C_1CB \iff \beta+\gamma+\theta-90^{\circ} = 90^{\circ}-\beta-\sigma \iff 2 \beta+ \gamma = 180^{\circ}-\sigma-\theta \]after some rearranging. But that is easy to prove. Let $C_M$ be the midpoint of the arc $\overarc{BC}$ not containing $C$. Then \[ 180^{\circ}-\sigma-\theta = \angle COC_M = 2 \angle CBC_M = 2 \left(\beta+\tfrac{\gamma}{2} \right). \]That last equality follows from simple angle chasing using that $CC_M$ is the angle bisector of $\angle ACB$. We're done, yay!

It suffices to prove that $ \angle OBA_2 = 90^{\circ} $ to show that $AA_2$ is a symmedian. For that, our similar triangles gives us $OA_1 * OA_2 = OA^2 = OB^2$. Specifically, we have: $OB^2 = OA_1 \cdot OA_2$. This can be rearranged to get: $\frac{OB}{OA_1} = \frac{OA_2}{OB}$ This, combined with the fact that $\angle BOA_1 = \angle BOA_2$, shows that triangles $BOA_1$ and $A_2OB$ are similar. Hence, $\angle OBA_2$ = 90 degrees as desired. Similarly, $\angle OCA_2$ = 90 degrees, and you have that $A_2B$ and $A_2C$ are tangent to (ABC). So $AA_2$ is a symmedian. Similarly, $BB_2$ and $CC_2$ are symmedians. Since it is well known that the symmedians of a triangle are concurrent, our three lines concur.

wait. does this even work? (My previous thing before the last edit did not work. I think the present version does work.)

vsathiam wrote: It suffices to prove that $ \angle OBA_2 = 90^{\circ} $ to show that $AA_2$ is a symmedian. For that, our similar triangles gives us that $OA_1 * OA_2 = OA^2 = OB^2$. So the power of $A_1$ with respect to the circle with diameter $A_1A_2$ = $OB^2 $, so OB is tangent to this circle. so $ \angle OBA_2 = 90^{\circ} $ as desired, and we can invoke the symmedians to get concurrency. Wouldn't it be the power of $O$, not $A_1$?

vsathiam wrote: wait. does this even work? I don't think $OB^2$ being the power of O with respect to $(A_1A_2)$ implies that it is tangent to $(A_1A_2)$. This would imply that any point of (ABC) is tangent.

ophiophagous wrote: Alternatively, you can draw the altitude $AH$. $AH \parallel OA_2$, so it's obvious that $\angle BAA_2 = \angle A_1AC$, so $AA_2$ is a symmedian, and similarly for $BB_2$, $CC_2$. I assume $H$ denotes the orthocenter. How does this imply that $\angle BAA_2 = \angle A_1AC$?

Delray wrote: vsathiam wrote: It suffices to prove that $ \angle OBA_2 = 90^{\circ} $ to show that $AA_2$ is a symmedian. For that, our similar triangles gives us that $OA_1 * OA_2 = OA^2 = OB^2$. So the power of $A_1$ with respect to the circle with diameter $A_1A_2$ = $OB^2 $, so OB is tangent to this circle. so $ \angle OBA_2 = 90^{\circ} $ as desired, and we can invoke the symmedians to get concurrency. Wouldn't it be the power of $O$, not $A_1$? Yeah it's the power of O. I'll fix that. Delray wrote: vsathiam wrote: wait. does this even work? I don't think $OB^2$ being the power of O with respect to $(A_1A_2)$ implies that it is tangent to $(A_1A_2)$. This would imply that any point of (ABC) is tangent. I agree, but the above symmedian logic still holds by similar triangles. Specifically, we have: $OB^2 = OA_1 \cdot OA_2$. This can be rearranged to get: $\frac{OB}{OA_1} = \frac{OA_2}{OB}$ This, combined with the fact that $\angle BOA_1 = \angle BOA_2$, shows that triangles $BOA_1$ and $A_2OB$ are similar. Hence, $\angle OBA_2$ = 90 degrees as desired. Similarly, $\angle OCA_2$ = 90 degrees, and you have that $A_2B$ and $A_2C$ are tangent to (ABC). So $AA_2$ is a symmedian. Similarly, $BB_2$ and $CC_2$ are symmedians. Since it is well known that the symmedians of a triangle are concurrent, our three lines concur.

We have $$\angle CAA_2=\angle OAA_2+\angle OAC=\angle OA_1A_2+(90^{\circ}-\angle B)=(90^{\circ}-\angle AA_1B)+(90^{\circ}-\angle B) = \angle BAA_1$$. Hence, $AA_2$ is symmedian. Similarly, $BB_2$ and $CC_2$ are symmedians. They concur at the symmedian point!

here is the generalization of above: assume triangle $ABC$ and point $P$ inside $ABC$ and let $D,E,F$ be the feet of altitudes from $P$ to $AB,AC,BC$ respectively . let $A',B',C'$ be points on the rays $PD,PE,PF$ respectively such that $PD.PA'=PE.PB'=PF.PC'$ . prove $AA',BB',CC'$ are concurrent

ophiophagous wrote: Alternatively, you can draw the altitude $AH$. $AH \parallel OA_2$, so it's obvious that $\angle BAA_2 = \angle A_1AC$, so $AA_2$ is a symmedian, and similarly for $BB_2$, $CC_2$. Are directed angles necessary for this solution?

Delray wrote: ophiophagous wrote: Alternatively, you can draw the altitude $AH$. $AH \parallel OA_2$, so it's obvious that $\angle BAA_2 = \angle A_1AC$, so $AA_2$ is a symmedian, and similarly for $BB_2$, $CC_2$. I assume $H$ denotes the orthocenter. How does this imply that $\angle BAA_2 = \angle A_1AC$? Using parallel lines and the fact that orthocentre is the isogonal conjugate of circumcentre

Solution : $\angle APB = \angle AA_1B + \angle A_1AA_2 = 90 - y$ $2 \angle B = \angle AOC = 2(90 + x - a + y)$ $\angle AOC = 180 - 2b$ $2(90 + x - a + y) = 180 - 2b \Longrightarrow a - x = b + y $ $\therefore AA_2$ is the $\text{Symmedian}$ of $\triangle ABC$. Similarly, $BB_2$ and $CC_2$ are $\text{Symmedians}$. Thus, $AA_2$, $BB_2$ and $CC_2$ concur at the $\text{ Lemoine Point of } \triangle ABC$

GMOH wrote: here is the generalization of above: assume triangle $ABC$ and point $P$ inside $ABC$ and let $D,E,F$ be the feet of altitudes from $P$ to $AB,AC,BC$ respectively . let $A',B',C'$ be points on the rays $PD,PE,PF$ respectively such that $PD.PA'=PE.PB'=PF.PC'$ . prove $AA',BB',CC'$ are concurrent I've been trying very hard to prove this, but then failed. Would you kindly plz gimme a hint? thanks.

WWY wrote: GMOH wrote: here is the generalization of above: assume triangle $ABC$ and point $P$ inside $ABC$ and let $D,E,F$ be the feet of altitudes from $P$ to $AB,AC,BC$ respectively . let $A',B',C'$ be points on the rays $PD,PE,PF$ respectively such that $PD.PA'=PE.PB'=PF.PC'$ . prove $AA',BB',CC'$ are concurrent I've been trying very hard to prove this, but then failed. Would you kindly plz gimme a hint? thanks. $ABC$ and $A'B'C'$ have common orthology center $P$

From similarity ratios we have $(OA_1)(OA_2)=OA^2=R^2$, so $(OA_1)(OA_1+A_1A_2)=R^2$. Thus, $(OA_1)(A_1A_2)=BA_1^2$, implying that $\triangle{OBA_2}$ is right. Similarly, we have $\triangle{OCA_2}$ is right, so $BA_2, CA_2$ are tangent to $(ABC)$, meaning that by tangent construction, $AA_2$ is the $A$-symmedian of $\triangle{ABC}$. Similarly, $BB_2$, $CC_2$ are also symmedians, and the three symmedians concur at the symmedian point, so we are done.

From the given condition of the problem we have $(OA_1)(OA_2)=OA^2$ and let let the tangent to $B,C$ to the circumcircle of $ABC$ meet at $A_2'$.This point lie on $OA_1$ and from $\triangle OA_1B\sim \triangle OBA_2'$ we have $(OA_1)(OA_2')=OB^2=OA^2$.Again note that $\angle A_1OA=\angle AOA_2'$ So $A_2'\equiv A_2$ and so $AA_2$ is the $A$-symmedian.

On dropping a perpendicular from $A$, such that it meets $BC$ at $D$, we observe that, since $AD \perp BC$ and $OA_1 \perp BC$ $\implies AD \parallel OA_1-(\bigstar)$. Also $\angle HAB=90^{\circ} - \angle B=\angle OAC$, so, if we prove that $\angle HAA_2 = \angle A_1AO$ then we would get: $\angle BAA_2 = \angle A_1AC$, which would ultimately imply that $AA_2$ is the symmedian. Now note that $\angle HAA_1=\angle OA_1A$ (from $\bigstar$), also $\angle A_2AO=\angle AA_1O$ (since $\triangle OAA_1 \sim \triangle OA_2A$) $\implies \angle HAA_2=\angle OAA_1$, which is the required condition, we wanted to prove. In a similar manner, we can prove that $BB_2$ and $CC_2$ are also symmedians, and as we know that the symmedians of a triangle concur, hence we're done. $\quad \blacksquare$

Let $\angle BAA_1 = \theta$. Then $\angle AA_1O = \angle BA_1O - \angle BA_1A = 90^\circ - (180^\circ - \theta - \angle B) = \angle B + \theta - 90^\circ$. But by the similarity condition, $\angle BAA_2 = \angle BAO - \angle A_2AO = \angle BAO - \angle AA_1O = (90^\circ - \angle C) - (\angle B + \theta - 90^\circ) = \angle A - \theta$ and hence $AA_2$ is the $A$-symmedian. But similarly $BB_2$ and $CC_2$ are the $B-$ and $C-$ symmedians, so the three lines concur about the symmedian point, as desired.

In fact, $\overline{AA_2},\overline{BB_2},\overline{CC_2}$ are the symmedians of $\triangle ABC$, which directly implies the concurrency, since the symmedians concur at the well-known symmedian point. I will show that $\overline{AA_2}$ is the $A$-symmedian, which implies that the other two are as well. Observe that from the triangle similarity, we get: $$\frac{OA}{OA_1}=\frac{OA_2}{OA} \implies OA_1\cdot OA_2=OB^2,$$as $OA=OB$. This implies that $\overline{OB}$ is tangent to $(A_1A_2B)$ by PoP, hence $\angle OBA_2=90^\circ$ and $A_2B$ is the tangent from $A_2$ to $(ABC)$. By symmetry, $A_2C$ is also a tangent. It is well-known that the line connecting $A$ and the intersection at the tangents at $B,C$ is the symmedian, so $\overline{AA_2}$ is the symmedian as desired. $\blacksquare$

It got bumped ..so.....

We claim that $AA_2$ is the $A$-symmedian, from which the desired concurrency is trivial. To show this, we will use Phantom Points. Let $A_2$ be the intersection of the tangents to the circumcircle at $B$ and $C$. We will show that $\triangle OAA_1\sim\triangle OA_2A$ in negative orientation. Toss on the complex plane with $(ABC)$ as the unit circle. Then $A=a,O=0,A_1=\frac{b+c}2,A_2=\frac{2bc}{b+c}$. Clearly $\measuredangle AOA_1=-\measuredangle A_2OA$ so it suffices to check that $\measuredangle OAA_1=-\measuredangle OA_2A$. Hence we just need to show $$\frac{a-0}{a-\frac{b+c}2}\cdot \frac{\frac{2bc}{b+c}-0}{\frac{2bc}{b+c}-a}\in\mathbb{R}.$$However, note \begin{align*}\frac{a-0}{a-\frac{b+c}2}\cdot \frac{\frac{2bc}{b+c}-0}{\frac{2bc}{b+c}-a}&= \frac{2a}{2a-b-c}\cdot\frac{2bc}{2bc-ab-ac}\\&= \frac{4abc}{4abc-2a^2b-2a^2c-2b^2c+ab^2+ac-2bc^2+abc+ac^2}\\&= \frac{4abc}{6abc-2abc\left(\frac{a}{c}+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}\right)+\frac{c}{b}+\frac{b}{c}}\\&= \frac{4}{6-2\left(\frac{a}{c}+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}\right)+\frac{c}{b}+\frac{b}{c}}\\&= \left(\overline{\frac{4}{6-2\left(\frac{a}{c}+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}\right)+\frac{c}{b}+\frac{b}{c}}}\right)\end{align*}so we are done!

Note that $OA^2=OA_1OA_2$, so $A_2$ is the inversion of $A_1$ with respect to the circumcircle. Claim: $A_2B$ and $A_2C$ are tangents to the circumcircle. We will use complex numbers with $(ABC)$ as the unit circle. Note that $A_1=\frac{b+c}{2}$. Then, $$A_2=\frac{1}{\overline{A_1}}=\frac{2}{\overline{b}+\overline{c}}=\frac{2}{1/b+1/c}=\frac{2bc}{b+c},$$so $A_2$ is the intersection of the tangents at $B$ and $C$. Thus, $AA_2$ is a symmedian, and similarly for the other vertices. Hence, they concur at the symmedian point, QED.

Claim: $AA_2$ is the $A$-Symmedian and so on. Proof. Note that the condition is equivalent to $OA_1 \cdot OA_2 = R^2$. As such, $\triangle OBA_1 \sim \triangle OBA_2$ so $A_2$ lies on the tangent to the circumcircle of $B$. By symmetry it also lies on the tangent at $C$. Thus, $AA_2$ is the $A$-Symmedian. $\blacksquare$ However, the symmedians simply concur at the isogonal conjugate to the median.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18.77051702864062cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.247718747332923, xmax = 14.522798281307695, ymin = -5.095299082902841, ymax = 4.471536379699366; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen ffqqff = rgb(1.,0.,1.); draw((2.,4.)--(1.,1.)--(5.490445091423507,0.944171418257164)--cycle, linewidth(2.)); /* draw figures */ draw((2.,4.)--(1.,1.), linewidth(2.) + zzttqq); draw((1.,1.)--(5.490445091423507,0.944171418257164), linewidth(2.) + zzttqq); draw((5.490445091423507,0.944171418257164)--(2.,4.), linewidth(2.) + zzttqq); draw(circle((3.2569375338486597,1.9143541553837802), 2.435120233820592), linewidth(2.) + linetype("4 4")); draw((xmin, -2.468340653957393*xmin + 3.468340653957393)--(xmax, -2.468340653957393*xmax + 3.468340653957393), linewidth(2.) + dotted); /* line */ draw((xmin, 2.302151411382367*xmin-11.695664498080856)--(xmax, 2.302151411382367*xmax-11.695664498080856), linewidth(2.) + dotted); /* line */ draw((2.,4.)--(3.1787088091421536,-4.37779552674069), linewidth(2.) + blue); draw((2.,4.)--(3.2569375338486597,1.9143541553837802), linewidth(2.) + green); draw((3.2569375338486597,1.9143541553837802)--(3.1787088091421536,-4.37779552674069), linewidth(2.) + ffqqff); draw((2.,4.)--(3.2452225457117536,0.9720857091285819), linewidth(2.) + blue); draw((2.,4.)--(1.9625529646796815,0.988032819514247), linewidth(2.) + green); /* dots and labels */ dot((2.,4.),dotstyle); label("$A$", (1.8578061386457059,4.142428160938436), NE * labelscalefactor); dot((1.,1.),dotstyle); label("$B$", (0.7456473304191154,0.8286488547938989), NE * labelscalefactor); dot((5.490445091423507,0.944171418257164),dotstyle); label("$C$", (5.625527815494972,0.8853916511319903), NE * labelscalefactor); dot((3.2452225457117536,0.9720857091285819),linewidth(4.pt) + dotstyle); label("$A_1$", (3.287724606365608,1.0669685994138827), NE * labelscalefactor); dot((3.7452225457117536,2.472085709128582),linewidth(4.pt) + dotstyle); label("$B_1$", (3.7870612141408118,2.564978422739495), NE * labelscalefactor); dot((1.5,2.5),linewidth(4.pt) + dotstyle); label("$C_1$", (1.5854407162228674,2.4628413893309307), NE * labelscalefactor); dot((3.2569375338486597,1.9143541553837802),linewidth(4.pt) + dotstyle); label("$O$", (3.299073165633226,2.0088990186261997), NE * labelscalefactor); dot((3.1787088091421536,-4.37779552674069),linewidth(4.pt) + dotstyle); label("$A_2$", (3.219633250759898,-4.289551374901944), NE * labelscalefactor); dot((1.9765700237261883,2.115463107489603),linewidth(4.pt) + dotstyle); label("$H$", (2.0166859683923617,2.2018245261757103), NE * labelscalefactor); dot((1.9625529646796813,0.988032819514247),linewidth(4.pt) + dotstyle); label("$D$", (2.0053374091247433,1.078317158681501), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Suppose the tangents of $B$ and $C$ to $\left(ABC\right)$ intersect at $A_2'$ (so that $AA_2'$ is the $A$-symmedian). Also let $H$ and $D$ be the orthocenter of $\Delta ABC$ and the foot from $A$ to $BC$ respectively. By proving $\measuredangle OA_1A = \measuredangle A_2AO$, we we can conclude that $A_2=A_2'$ and hence prove the collinearity (as they meet at the symmedian point). We proceed by angle chasing: $\measuredangle OA_1A = \measuredangle OA_1D - \measuredangle AA_1D = \frac{\pi}{2} - \measuredangle AA_1D$ and $\measuredangle A_2AO = \measuredangle DAA_1 = \frac{\pi}{2} - \measuredangle AA_1D$, where the first inequality is due to the isogonality of the green and blue lines as highlighted in the figure. We are done.

Our similar triangle ratios give $OA_1 \cdot OA_2 = OA^2$ and since $A_2$ lies on ray $OA_1$ it follows that $A_2$ is the inverse of $A_1$ wrt $(ABC)$. However since $A_1$ is the midpoint of $BC$, $A_2$ is the intersection of the tangents at $B$ and $C$ so $AA_2$ is a symmedian, similarly for $BB_2$ and $CC_2$ so we are done(concurrency point is symmedian point).

Notice that $A_1$, $A_2$ are inverses with respect to $(ABC)$, as our similarity gives \[OA_1 \cdot OA_2 = OA^2 = R^2.\] Thus $A_2$ is the intersection of the tangents at $B$ and $C$, so $AA_2$ is the $A$-symmedian. Hence our desired concurrency point is simply the symmedian point. $\blacksquare$ [asy][asy] size(250); pair O, A, B, C, A1, A2; O = (0,0); A = dir(120); B = dir(210); C = dir(330); A1 = (B+C)/2; A2 = -2/(B+C); draw(A--B--C--cycle^^A--O--A2--A--A1); draw(B--A2--C, dashed); draw(circumcircle(A, B, C)); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$A_1$", A1, dir(315)); dot("$A_2$", A2, dir(315)); dot("$O$", O, dir(0)); [/asy][/asy]