$\triangle ABC$ with $AB=AC$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -7.3, xmax = 12.26, ymin = -6.72, ymax = 4.72; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((-0.84,-1.32)--(3.16,-0.34)--(0.78,1.56)--cycle, linewidth(2) + zzttqq); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1, gridy = 1; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((-0.84,-1.32)--(3.16,-0.34), linewidth(2) + zzttqq); draw((3.16,-0.34)--(0.78,1.56), linewidth(2) + zzttqq); draw((0.78,1.56)--(-0.84,-1.32), linewidth(2) + zzttqq); draw(circle((1.0802573396158028,-0.5045197535338892), 2.086239747182665), linewidth(2)); draw((xmin, 0.7591588809435373*xmin-0.6823065400074286)--(xmax, 0.7591588809435373*xmax-0.6823065400074286), linewidth(2)); /* line */ draw(circle((0.49975463915983515,-0.302913367396507), 1.6820843353919606), linewidth(2)); /* dots and labels */ dot((-0.84,-1.32),dotstyle); label("$A$", (-0.76,-1.12), NE * labelscalefactor); dot((3.16,-0.34),dotstyle); label("$B$", (3.24,-0.14), NE * labelscalefactor); dot((0.78,1.56),dotstyle); label("$C$", (0.86,1.76), NE * labelscalefactor); label("$c$", (1.22,-1.14), NE * labelscalefactor,zzttqq); label("$a$", (2.16,0.88), NE * labelscalefactor,zzttqq); label("$b$", (-0.32,0.28), NE * labelscalefactor,zzttqq); label("$d$", (0.02,0.88), NE * labelscalefactor); label("$f$", (-7.2,-5.9), NE * labelscalefactor); dot((1.8395092783196703,0.7141732652069861),linewidth(4pt) + dotstyle); label("$D$", (1.92,0.88), NE * labelscalefactor); label("$e$", (-0.36,0.74), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Since both circles, by definition, pass through $A$, don't they automatically have to intersect? Sorry that I'm missing something here!

lol yes @above is correct they meet at $A$ by definition! Also, you need to specify bisector. Angle bisector? Perpendicular bisector?

@2above touch is equivalent to tangent @above it's pretty clear $AD$ refers to the angle bisector The problem is not difficult: since the tangent to $k$ at $D$ is parallel to the tangent to $(ABC)$ at the arc midpoint $M = AD \cap (ABC)$, $A$ is the center of homothety sending $k$ to $(ABC)$.

@cw357, circle $k$ must be tangent to BC at D. This is what is implied by "touches," if it wasn't clear.

No; my point was that since the problem literally states that $k$ passes through $A$, and the circumcircle passes through $A$, of course the circumcircle meets $k$ at point $A$. What is wrong with that solution? (I'm assuming that something is wrong because that seemed a little too simple for an HSM problem. XD)

Maybe a translation error

cw357 wrote: No; my point was that since the problem literally states that $k$ passes through $A$, and the circumcircle passes through $A$, of course the circumcircle meets $k$ at point $A$. What is wrong with that solution? (I'm assuming that something is wrong because that seemed a little too simple for an HSM problem. XD) I don't see anything wrong with the problem statement? It asks you to prove $(ABC)$ is tangent to $k$ at $A$, not just intersect.

MP8148 wrote: cw357 wrote: No; my point was that since the problem literally states that $k$ passes through $A$, and the circumcircle passes through $A$, of course the circumcircle meets $k$ at point $A$. What is wrong with that solution? (I'm assuming that something is wrong because that seemed a little too simple for an HSM problem. XD) I don't see anything wrong with the problem statement? It asks you to prove $(ABC)$ is tangent to $k$ at $A$, not just intersect. The circumcircle is intersecting the vertex, not a side, so saying "tangent" isn't really applicable here.

cw357 wrote: No; my point was that since the problem literally states that $k$ passes through $A$, and the circumcircle passes through $A$, of course the circumcircle meets $k$ at point $A$. What is wrong with that solution? (I'm assuming that something is wrong because that seemed a little too simple for an HSM problem. XD) In both uses, I believe touches is implied as tangent. So $k$ must be tangent to the circumcircle and it must be tangent to $BC$ at $D$. In the diagram you have provided, circle k is not tangent to $BC$, so your diagram is wrong.

natmath wrote: cw357 wrote: No; my point was that since the problem literally states that $k$ passes through $A$, and the circumcircle passes through $A$, of course the circumcircle meets $k$ at point $A$. What is wrong with that solution? (I'm assuming that something is wrong because that seemed a little too simple for an HSM problem. XD) In both uses, I believe touches is implied as tangent. So $k$ must be tangent to the circumcircle and it must be tangent to $BC$ at $D$. In the diagram you have provided, circle k is not tangent to $BC$, so your diagram is wrong. Actually, I made the diagram on Geogebra specifically specifying the angle bisector and making sure that $k$ was tangent to $D$ (try it yourself if you want), so I'm fairly sure that parmenides51 wrote: touches does not necessarily imply tangent. EDIT: hold on just realized I messed up the geogebra diagram will remake it and share by tomorrow morning

touches means tangent, otherwise it should have 2 letters for 2 points if it implied intersect

parmenides51 wrote: touches means tangent, otherwise it should have 2 letters for 2 points if it implied intersect Thank you so much for the clarification! Revised diagram below shows that both circles are definitely tangent at $A$; sorry about my earlier mistake! [asy][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 = -5.382222222222224, xmax = 6.688888888888899, ymin = -1.8355555555555565, ymax = 3.248888888888889; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((-1.06,-0.78)--(3,0)--(-1.14,2.06)--cycle, linewidth(2) + zzttqq); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 0.5, gridy = 0.5; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 0.5, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 0.5, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((-1.06,-0.78)--(3,0), linewidth(2) + zzttqq); draw((3,0)--(-1.14,2.06), linewidth(2) + zzttqq); draw((-1.14,2.06)--(-1.06,-0.78), linewidth(2) + zzttqq); draw((xmin, 1.2457202181433884*xmin + 0.5404634312319919)--(xmax, 1.2457202181433884*xmax + 0.5404634312319919), linewidth(2)); /* line */ draw((xmin, 2.009708737864077*xmin + 0.123130334276188)--(xmax, 2.009708737864077*xmax + 0.123130334276188), linewidth(2)); /* line */ draw((xmin, -0.8027484706721641*xmin + 0.014268892790712936)--(xmax, -0.8027484706721641*xmax + 0.014268892790712936), linewidth(2)); /* line */ draw(circle((-0.03870687922115358,0.045340780889986144), 1.3130982610415292), linewidth(2)); draw(circle((0.7620412669933061,0.6924518666758677), 2.342637163178246), linewidth(2)); /* dots and labels */ dot((-1.06,-0.78),dotstyle); label("$A$", (-1.0266666666666646,-0.688888888888889), NE * labelscalefactor); dot((3,0),dotstyle); label("$B$", (3.0355555555555616,0.08444444444444446), NE * labelscalefactor); dot((-1.14,2.06),dotstyle); label("$C$", (-1.1066666666666647,2.146666666666667), NE * labelscalefactor); label("$c$", (0.9911111111111152,-0.528888888888889), NE * labelscalefactor,zzttqq); label("$a$", (0.9911111111111152,1.16), NE * labelscalefactor,zzttqq); label("$b$", (-1.2488888888888872,0.6444444444444446), NE * labelscalefactor,zzttqq); label("$f$", (1.96,3.1066666666666674), NE * labelscalefactor); dot((0.5462557174398107,1.220945222723186),linewidth(4pt) + dotstyle); label("$D$", (0.5822222222222259,1.2933333333333337), NE * labelscalefactor); label("$g$", (1.3644444444444488,3.1066666666666674), NE * labelscalefactor); label("$h$", (-3.853333333333334,3.1066666666666674), NE * labelscalefactor); dot((-0.03870687922115358,0.045340780889986144),linewidth(4pt) + dotstyle); label("$E$", (-0.004444444444441417,0.12), NE * labelscalefactor); label("$d$", (-0.6977777777777754,0.9911111111111113), NE * labelscalefactor); label("$e$", (-0.4133333333333307,2.5111111111111115), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]but it still looks like you can use homothety here; good problem!