Tangent in Isosceles Triangle

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Source: All Russian Olympiad 2017 10.2 & 11.2

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#1 h Jul 5, 2017, 8:45 PM • 3 Y

Y by donotoven, Adventure10, Mango247

Let $ABC$ be an acute angled isosceles triangle with $AB=AC$ and circumcentre $O$ . Lines $BO$ and $CO$ intersect $AC, AB$ respectively at $B', C'$ . A straight line $l$ is drawn through $C'$ parallel to $AC$ . Prove that the line $l$ is tangent to the circumcircle of $\triangle B'OC$ .

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#3 h May 13, 2018, 6:38 PM • 4 Y

Y by AforApple, donotoven, Adventure10, Mango247

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(16cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.52022599245634, xmax = 3.116666969996216, ymin = -6.643620687561152, ymax = 5.644711030990139; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-4.46,4.87)--(-8.018898138313551,-4.352121533938173), linewidth(2) + wrwrwr); draw((-8.018898138313551,-4.352121533938173)--(-0.39656158458527013,-4.141201351082423), linewidth(2) + wrwrwr); draw((-0.39656158458527013,-4.141201351082423)--(-4.46,4.87), linewidth(2) + wrwrwr); draw(circle((-4.31182143610446,-0.48494929298616624), 5.356999049584682), linewidth(2) + wrwrwr); draw((xmin, 1.0431864651323544*xmin + 4.013084469225557)--(xmax, 1.0431864651323544*xmax + 4.013084469225557), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.9338465891804261*xmin-4.511529034247363)--(xmax, -0.9338465891804261*xmax-4.511529034247363), linewidth(2) + wrwrwr); /* line */ draw((xmin, -2.2176296106514775*xmin-12.136973512968279)--(xmax, -2.2176296106514775*xmax-12.136973512968279), linewidth(2) + red); /* line */ draw((xmin, -36.13848826852717*xmin-156.30765767763117)--(xmax, -36.13848826852717*xmax-156.30765767763117), linewidth(2) + wrwrwr); /* line */ draw(circle((-1.617853472730625,-1.52457531984392), 2.887608987277295), linewidth(2) + linetype("4 4") + blue); /* dots and labels */ dot((-4.46,4.87),dotstyle); label("$A$", (-4.406101811899774,5.001483198394216), NE * labelscalefactor); dot((-8.018898138313551,-4.352121533938173),dotstyle); label("$B$", (-7.970656050868851,-4.21811573548069), NE * labelscalefactor); dot((-0.39656158458527013,-4.141201351082423),dotstyle); label("$C$", (-0.3457261186380053,-4.003706457948716), NE * labelscalefactor); dot((-4.31182143610446,-0.48494929298616624),linewidth(4pt) + dotstyle); label("$O$", (-4.258695433596541,-0.37214931975089627), NE * labelscalefactor); dot((-2.770383953826538,1.123057425373856),linewidth(4pt) + dotstyle); label("$B'$", (-2.717628751335474,1.235920261738913), NE * labelscalefactor); dot((-5.939823436816552,1.0353548225577307),linewidth(4pt) + dotstyle); label("$C'$", (-5.880165594932099,1.142116202818674), NE * labelscalefactor); dot((-4.250207390643089,-2.7115877520684135),linewidth(4pt) + dotstyle); label("$X$", (-4.1916925343677995,-2.6100461539908806), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let the intersection between $AO$ and $l$ be $X$ . Since the triangle is isosceles, $XB'=XC'$ , and $AB'=AC'$ so $AB'XC'$ is a rhombus.

Let $\angle BAC=\theta$ .

Note that $\angle XOC=\theta$ and $\angle XB'C=180^{\circ}-XB'A=\theta$ , so $OB'CX$ is a circle.

Now, $\angle C'XB'=\theta=\angle AOB'=\angle ACX$ , so $C'X$ , or $l$ , is tangent to $(B'OC)$ , as desired.

This post has been edited 1 time. Last edited by lifeisgood03, May 13, 2018, 6:38 PM

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#4 h May 26, 2018, 6:25 AM • 5 Y

Y by socr4tes, donotoven, third_one_is_jerk, Adventure10, Mango247

We prove a stronger result, all because of the amazing symmetry!

utkarshgupta wrote:

Let $ABC$ be an acute-angled isosceles triangle with $AB=AC$ and circumcentre $O$ . Lines $BO$ and $CO$ intersect $AC, AB$ respectively at $B', C'$ . A straight line $l$ is drawn through $C'$ parallel to $AC$ . Prove that the line $l$ is tangent to the circumcircle of $\triangle B'OC$ .

Let $M$ be the second intersection on $\omega_1=\odot(B'OC)$ and $\omega_2=\odot(C'OB)$ . We will show that $C'M$ is both tangent to $\omega_1$ and parallel to $AC$ . Clearly, $A, O, M$ are collinear, as $A$ lies on the radical axis of $\omega_1, \omega_2$ .

Note the perfect symmetry about $AM$ . This yields $\angle C'MA=\angle AMB' \overset{\omega_1}{=} \angle OCB'=\angle OCA=\tfrac{\pi}{2}-\angle B=\angle CAO$ , and so $C'M \parallel AC$ . It suffices to show $C'M$ is tangent to $\omega_1$ . This is further equivalent to $\angle B'CM=\angle B'MC'=2\left(\tfrac{\pi}{2}-\angle B\right)=\pi-2\angle B$ . But this follows since $\angle B'CM \overset{\omega_1}{=} \pi-\angle B'OM=\pi-2\angle C=\pi-2\angle B$ , as desired. $\square$

Comments:
In fact, $O$ is the circumcenter of $\triangle MB'C'$ also!
Proof: Since the sides of the triangle $MB'C'$ are respectively parallel to the sides of $\triangle ABC$ , hence these triangles are homothetic. Also, since $O=AM' \cap CC' \cap BB'$ , hence $O$ is the center of homothety. But $O$ is the circumcenter of $\triangle ABC$ , and hence we get that $O$ is the circumcenter of $\triangle MB'C'$ as well. $\square$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.037904761904763, xmax = 12.651154471544714, ymin = -10.22185133565621, ymax = 7.429252032520318; /* image dimensions */pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */draw(circle((-2.78,-0.25), 6.440279497040482), linewidth(1) + wrwrwr); draw((-7.6796928721752185,-4.429738001161718)--(2.19671889463137,-4.337721742464758), linewidth(1) + wrwrwr); draw((2.19671889463137,-4.337721742464758)--(-2.84,6.19), linewidth(1) + wrwrwr); draw((-2.84,6.19)--(-7.6796928721752185,-4.429738001161718), linewidth(1) + wrwrwr); draw(circle((0.2521360268231692,-1.6318238373561076), 3.3321593603314583), linewidth(1) + wrwrwr); draw(circle((-5.785863644866286,-1.6880784926824097), 3.332159360331457), linewidth(1) + wrwrwr); draw((-7.6796928721752185,-4.429738001161718)--(-2.7537276180431176,-3.0699023300385186), linewidth(1) + wrwrwr); draw((-2.7537276180431176,-3.0699023300385186)--(2.19671889463137,-4.337721742464758), linewidth(1) + wrwrwr); draw((-2.78,-0.25)--(-2.7537276180431176,-3.0699023300385186), linewidth(1) + wrwrwr); draw((-4.959170994862998,1.539903115857993)--(-2.78,-0.25), linewidth(1) + wrwrwr); draw((-2.78,-0.25)--(-0.6345566231801201,1.5801945541034892), linewidth(1) + wrwrwr); draw((-2.78,-0.25)--(2.19671889463137,-4.337721742464758), linewidth(1) + wrwrwr); draw((-2.78,-0.25)--(-7.6796928721752185,-4.429738001161718), linewidth(1) + wrwrwr); draw((-4.959170994862998,1.539903115857993)--(-2.7537276180431176,-3.0699023300385186), linewidth(1) + wrwrwr); draw((-2.7537276180431176,-3.0699023300385186)--(-0.6345566231801201,1.5801945541034892), linewidth(1) + wrwrwr); draw((-4.959170994862998,1.539903115857993)--(-0.6345566231801201,1.5801945541034892), linewidth(1) + wrwrwr); /* dots and labels */dot((-2.78,-0.25),dotstyle); label("$O$", (-3.1087990708478526,0.24395586527293572), NE * labelscalefactor); dot((-2.84,6.19),dotstyle); label("$A$", (-2.998740998838561,6.732202090592327), NE * labelscalefactor); dot((-7.6796928721752185,-4.429738001161718),dotstyle); label("$B$", (-8.392887340301976,-4.982734030197444), NE * labelscalefactor); dot((2.19671889463137,-4.337721742464758),linewidth(4pt) + dotstyle); label("$C$", (2.3102020905923332,-4.982734030197444), NE * labelscalefactor); dot((-0.6345566231801201,1.5801945541034892),linewidth(4pt) + dotstyle); label("$B'$", (-0.547823461091755,1.7629105691056872), NE * labelscalefactor); dot((-4.959170994862998,1.539903115857993),linewidth(4pt) + dotstyle); label("$C'$", (-5.652027874564461,1.9652799070847813), NE * labelscalefactor); label("$w_1$", (-2.3016910569105704,1.1108315911730513), NE * labelscalefactor,wrwrwr); label("$w_2$", (-7.968032520325204,1.4481138211382079), NE * labelscalefactor,wrwrwr); dot((-2.7537276180431176,-3.0699023300385186),linewidth(4pt) + dotstyle); label("$M$", (-3.186313588850175,-4.215858304297329), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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#5 h Jan 22, 2021, 10:01 AM • 1 Y

Y by donotoven

First of all , forget about the tangents and the circles , draw a line parallel to $AB$ from $B'$ and a line parallel to $AC$ from $C'$ and let the lines intersect in a point $X$ , we make the following claim:
$\text Claim$ : the point of tangency of $(B'OC)$ and parallel to $AC$ from $C'$ is $X$ .
First , we’ll show that $X$ lies on both $(B'OC)$ and $(C'OB)$ . By symmetry we can just show that $X$ lies on $(B'OC)$ , note that because $\angle C'BC = \angle B'CB = 90 - \frac{\angle A}{2}$ and $\angle C'CB = \angle B'BC = 90 - \angle A$ , $\triangle C'BC = \triangle B'CB$ thus $C'B=B'C$ and $AC'=AB'$ and because $AB'XC'$ was a parallelogram , its a Rhombus. Now note that $C'X = B'X$ , $C'B = B'C$ , $\angle BC'X = \angle CB'X = \angle A$ therefore $\triangle BXC' = \triangle CXB'$ and $XB = XC$ and so $X$ lies on the perpendicular bisector of $BC$ and so $A , O , X$ are collinear .
Now in order to prove that $B'OXC$ and $C'OXB$ are cyclic we should just show that $\angle B'CO = \angle B'XO$ .
Since $A , O , X$ are collinear , $\angle B'XO = \angle B'XA$ and since $B'X \parallel AB$ , $\angle B'XA = \angle OAB$ and because $OA = OB$ , $\angle OAB = \angle OBA = \angle OBC'$ and thus $B'OXC$ is cyclic , similarly $C'OXB$ is cyclic.
Now we just need to show that $C'X$ is tangent to $(B'OXC)$ which is equivalent to $\angle OB'X = \angle OXC'$ , we knew that $B'X \parallel AB$ , so $\angle OB'X = \angle OBC'$ and because $C'OXB$ Is cyclic , $\angle C'BO = \angle C'XO$ therefore $\angle OB'X = \angle OBC' = \angle C'XO$ and we are Done

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#6 h Jul 6, 2021, 4:41 PM • 1 Y

Y by donotoven

Let $AO$ intersect circle $B'OC$ at $K (K \ne B)$ . Now we prove that $O$ is the circumcenter of $KB'C'$ and $C'K$ is tangent to circle $B'OC$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = 9.965242981317132, xmax = 10.20565007198979, ymin = 2.638169113776028, ymax = 2.7699430538506142; /* image dimensions */ pen ttqqqq = rgb(0.2,0.,0.); pen qqzzcc = rgb(0.,0.6,0.8); pen ffwwqq = rgb(1.,0.4,0.); pen dcrutc = rgb(0.8627450980392157,0.0784313725490196,0.23529411764705882); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen qqzzff = rgb(0.,0.6,1.); draw(arc((10.078328942637937,2.7122371116861053),0.00482099780760678,90.14209168102435,124.05721905955254)--(10.078328942637937,2.7122371116861053)--cycle, linewidth(0.8) + qqwuqq); draw(arc((10.078328942637937,2.7122371116861053),0.00482099780760678,56.22696430248275,90.14209168102435)--(10.078328942637937,2.7122371116861053)--cycle, linewidth(0.8) + qqwuqq); draw(arc((10.078328942637937,2.7122371116861053),0.00482099780760678,-89.8579083189641,-55.94278094055423)--(10.078328942637937,2.7122371116861053)--cycle, linewidth(0.8) + qqwuqq); draw(arc((10.088195589414552,2.726990747847996),0.00482099780760678,-106.81547200818294,-72.90034462975353)--(10.088195589414552,2.726990747847996)--cycle, linewidth(0.8) + qqwuqq); draw(arc((10.078211870773865,2.7594440270473353),0.00482099780760678,-89.85790831897565,-72.90034462978403)--(10.078211870773865,2.7594440270473353)--cycle, linewidth(0.8) + qqzzff); draw(arc((10.078211870773865,2.7594440270473353),0.00482099780760678,-106.81547200822092,-89.85790831897566)--(10.078211870773865,2.7594440270473353)--cycle, linewidth(0.8) + qqzzff); draw(arc((10.104765866731599,2.6731270668355864),0.00482099780760678,107.0996553702465,124.0572190594458)--(10.104765866731599,2.6731270668355864)--cycle, linewidth(0.8) + qqzzff); draw(arc((10.078372959084787,2.6944883494428242),0.00482099780760678,73.18452799181709,90.14209168103591)--(10.078372959084787,2.6944883494428242)--cycle, linewidth(0.8) + qqzzff); /* draw figures */ draw((10.052086326228965,2.6729964230102516)--(10.078211870773865,2.7594440270473353), linewidth(0.4)); draw((10.078211870773865,2.7594440270473353)--(10.104765866731599,2.6731270668355864), linewidth(0.4)); draw((10.104765866731599,2.6731270668355864)--(10.052086326228965,2.6729964230102516), linewidth(0.4)); draw(circle((10.078328942637938,2.712237111686105), 0.047207060528524046), linewidth(0.4) + ttqqqq); draw(circle((10.107454822550814,2.7034349073733845), 0.03042689076232164), linewidth(0.4) + qqzzcc); draw((10.078211870773865,2.7594440270473353)--(10.078372959084787,2.6944883494428242), linewidth(0.4)); draw((10.078328942637937,2.7122371116861053)--(10.104765866731599,2.6731270668355864), linewidth(0.4)); draw((10.078372959084787,2.6944883494428242)--(10.088195589414552,2.726990747847996), linewidth(0.4)); draw((10.068389240444075,2.726941628642159)--(10.078328942637937,2.7122371116861053), linewidth(0.4) + ffwwqq); draw((10.078328942637937,2.7122371116861053)--(10.088195589414552,2.726990747847996), linewidth(0.4) + ffwwqq); draw(circle((10.078328942637937,2.7122371116861053), 0.01774881682299166), linewidth(0.4) + linetype("4 4") + dcrutc); /* dots and labels */ dot((10.078211870773865,2.7594440270473353),linewidth(2.pt) + dotstyle); label("$A$", (10.078857829649733,2.7601403583084805), NE * labelscalefactor); dot((10.052086326228965,2.6729964230102516),linewidth(2.pt) + dotstyle); label("$B$", (10.052663741561735,2.673683797625398), NE * labelscalefactor); dot((10.104765866731599,2.6731270668355864),linewidth(2.pt) + dotstyle); label("$C$", (10.10537331759157,2.6738444975523183), NE * labelscalefactor); dot((10.078328942637937,2.7122371116861053),linewidth(2.pt) + dotstyle); label("$O$", (10.079500629357414,2.7116089803785717), NE * labelscalefactor); dot((10.088195589414552,2.726990747847996),linewidth(2.pt) + dotstyle); label("$B'$", (10.088821225118787,2.7276789730705944), NE * labelscalefactor); dot((10.068389240444075,2.726941628642159),linewidth(2.pt) + dotstyle); label("$C'$", (10.065841135569194,2.7288038725590362), NE * labelscalefactor); dot((10.078372959084787,2.6944883494428242),linewidth(2.pt) + dotstyle); label("$K$", (10.079018529576652,2.6950568879057886), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
$\textbf{Claim: }$ $KO= OB'$ .
$\textbf{Proof: }$ Notice that, $\measuredangle CB'O = \measuredangle CAO + \measuredangle OKB'$ , Now $\measuredangle AOB' = \measuredangle COK = \measuredangle CB'K$ $\therefore \measuredangle KB'O = \measuredangle B'AO = \measuredangle OCB' = \measuredangle OKB'$ . $\blacksquare$

Because of symmetry, $OB'= OC'$ . For the claim we get that $O$ is the circumcenter of $OC'B'$ .
Since $K$ lies on the perpendicular bisector of $B'C'$ , hence, $B'K = C'K \implies \measuredangle KB'O= \measuredangle OKB' = \measuredangle C'KO$ .

So, $C'K$ is tangent to the circle $B'CO$ . And we are done $\blacksquare$

This post has been edited 2 times. Last edited by Math-Star., Jul 6, 2021, 4:43 PM

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#7 h Jul 15, 2021, 5:59 PM • 1 Y

Y by donotoven

Let $D = AO \cap l$ .
$\angle DC'C = \angle C'CA = \angle OCA = \angle OAC = \angle DAC$ So $DC'AC$ is an isosceles trapezium.
$\angle OB'C = \angle OC'B = 180^\circ - \angle OC'A = 180^\circ - \angle ODC$ So $ODCB'$ is cyclic.
$\angle OC'D = \angle OCB' = \angle OAC'$ So $(AOC')$ is tangent to $l$ at $C'$ .
$C'D^2 = DO \times DA = C'O \times C'C$ So $(OB'C)$ is tangent to $l$ at $D$ .

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#8 h Aug 29, 2021, 7:48 PM

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Let $A'=AO\cap \ell.$ Easy to note that trapezoid $AC'A'C$ is isosceles, so $\angle C'A'O=\angle B'CO=\angle OAB'=\angle OAC'=\angle OCA',$ $|A'O|=|OC'|=|OB'|.$ Hence $A'\in (B'OC),$ and $\ell$ tangent to $(B'OC)$ at $A'$ . Done.

This post has been edited 2 times. Last edited by JAnatolGT_00, Aug 29, 2021, 7:54 PM

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#9 h Aug 29, 2021, 9:10 PM

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Let $D=AO\cap (B'OC)$ . Note that $\measuredangle DB'C=\measuredangle DOC=\measuredangle BOD=\measuredangle B'OA=\measuredangle B'OD=\measuredangle B'CD,$ thus $\triangle CDB'$ is isosceles. Now, $\measuredangle B'CD=\measuredangle DOC=2\measuredangle ACO$ , hence $CO$ is the angle bisector of $\angle B'CD$ , meaning $OC'=OB'=OD$ .
Now, we conclude that $\measuredangle DC'O=\measuredangle ODC'=\measuredangle ACO=\measuredangle OCD$ , therefore $C'D$ is parallel to $AC$ and $CD'$ is tangent to $(B'OC)$ .

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third_one_is_jerk

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third_one_is_jerk

#10 h Sep 24, 2021, 1:18 PM

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For storage.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.879454953348385, xmax = 2.2140922767978015, ymin = -1.295154732330985, ymax = 1.168455284767441; /* image dimensions */ /* draw figures */ draw(circle((0,0), 1), linewidth(0.8)); draw((-0.6,-0.8)--(0.6,-0.8), linewidth(0.8)); draw((0,1)--(-0.6,-0.8), linewidth(0.8)); draw((0,1)--(0.6,-0.8), linewidth(0.8)); draw(circle((0.576923076923077,-0.19230769230769224), 0.6081303192631499), linewidth(0.8)); draw(circle((0,-0.3846153846153844), 0.72975638311578), linewidth(0.8)); draw((0,1)--(0,-0.3846153846153844), linewidth(0.8)); draw((-0.23076923076923078,0.3076923076923077)--(0,-0.3846153846153844), linewidth(0.8)); draw((-0.6,-0.8)--(0.23076923076923064,0.30769230769230765), linewidth(0.8)); draw((0.6,-0.8)--(-0.23076923076923078,0.3076923076923077), linewidth(0.8)); /* dots and labels */ dot((-0.6,-0.8),dotstyle); label("$B$", (-0.6957464527251732,-0.9270225278324325), NE * labelscalefactor); dot((0.6,-0.8),dotstyle); label("$C$", (0.6082033177130376,-0.9296945560505436), NE * labelscalefactor); dot((0,1),linewidth(4pt) + dotstyle); label("$A$", (0.009668996856153977,1.021493732771331), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.049184968717866,0), NE * labelscalefactor); dot((0.23076923076923064,0.30769230769230765),linewidth(4pt) + dotstyle); label("$B'$", (0.2094634236137075,0.3668468193341136), NE * labelscalefactor); dot((-0.23076923076923078,0.3076923076923077),linewidth(4pt) + dotstyle); label("$C'$", (-0.33502264328017634,0.3695188475522247), NE * labelscalefactor); dot((0,-0.3846153846153844),linewidth(4pt) + dotstyle); label("$X$", (0.05,-0.3813210817369919), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $X= \odot(OB'C) \cap OA.$ Notice that $X \in \odot (OC'B)$ as well. By spiral similarity lemma, we have $\triangle XBC' \overset{+}{\sim} \triangle XB'C$ .But $\triangle XBC' \overset{-}{\sim} \triangle XCB'$ and $XC = XB$ by symmetry. Therefore, $X$ is the center of $\odot (BCB'C)$ . Since $AO$ bisects $\overline{B'C'}$ ,
$\begin{align*} \angle C'XA = \frac{1}{2}\angle C'XB' =&\hspace{1mm} \angle C'BB' = \angle ABO = \angle OAB = \angle XAC \\ &\implies X \text{ lies on } l. \end{align*}$ We are done if $\angle XB'O = \frac{\angle A}{2}.$ Indeed, by symmetry $XB'$ is parallel to $AB$ and $\angle XB'O = \angle XAB. \blacksquare$

This post has been edited 1 time. Last edited by third_one_is_jerk, Sep 24, 2021, 1:20 PM

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