From the tangency we have $\angle ABD=\angle BDX=\angle CDX$, where X is the intersection of the angle bisector of $\angle ADC$ with $AC$. So $AB$ and $DX$ are parallel, so $\angle BAD=\angle ADX$ so $ABD$ is an isoscel triangle, so $BD=DA=DC$, so $\angle A=90$.

This was a very simple angle chasing problem.

Let the intersection point of AC and tangent of circumcircle of ABD be X,and angle ADX=y{say} =>angle XDC=y By alternate segment theorem angle DAB=y, angle DAB+DBA=ADC=2y Angle DBA=y =>DA=DC,Angle DAX=DCX=90-y,as angle ADC=2y,sum of angles in triangle ADC=180 Angle A=BAD+DAX=y+90-y=90

Nice to see that 4 months ago, I wasn't able to solve this problem, now I can do it MENTALLY Let $\ell$ be the angle bisector of $\angle ADC$, Let $X \in \ell$ such that, $AX>BX$, and let $\ell \cap AC=Y$, then, $\angle ABC=\angle ADY$ $=$ $\angle YDC$ $=$ $\angle XDB$ $=$ $\angle DAB$ $\implies$ $BD=CD=AD$ $\implies$ $\angle A=90^{\circ}$

Proof without words [asy][asy] 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 = -5.961720780579867, xmax = 7.281118273712037, ymin = -2.755441953669829, ymax = 5.037669094740618; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-1.007231723236255,3.4887187239849142)--(-3.0556368653973305,-0.26090424810654206)--(3.558281432597328,-0.2782636137180766)--cycle, linewidth(0.8)); draw(arc((0.25132228359999886,-0.26958393091230937),0.4087296004411082,54.181961764721294,108.51430580751418)--(0.25132228359999886,-0.26958393091230937)--cycle, linewidth(2) + qqwuqq); draw(arc((0.25132228359999886,-0.26958393091230937),0.4087296004411082,-0.15038227807159213,54.181961764721294)--(0.25132228359999886,-0.26958393091230937)--cycle, linewidth(2) + ffvvqq); draw(arc((-3.0556368653973305,-0.26090424810654206),0.4087296004411082,-0.1503822780715931,61.35238354319109)--(-3.0556368653973305,-0.26090424810654206)--cycle, linewidth(2) + qqwuqq); draw(arc((-1.007231723236255,3.4887187239849142),0.4087296004411082,-118.6476164568089,-71.48569419248584)--(-1.007231723236255,3.4887187239849142)--cycle, linewidth(2) + ffvvqq); draw(arc((-1.007231723236255,3.4887187239849142),0.4087296004411082,-71.48569419248584,-39.52587179939077)--(-1.007231723236255,3.4887187239849142)--cycle, linewidth(2) + blue); draw(arc((3.558281432597328,-0.2782636137180766),0.4087296004411082,140.47412820060924,179.8496177219284)--(3.558281432597328,-0.2782636137180766)--cycle, linewidth(2) + blue); draw(arc((0.25132228359999886,-0.26958393091230937),0.4087296004411082,179.8496177219284,234.1819617647213)--(0.25132228359999886,-0.26958393091230937)--cycle, linewidth(2) + ffvvqq); /* draw figures */ draw((-1.007231723236255,3.4887187239849142)--(-3.0556368653973305,-0.26090424810654206), linewidth(0.8)); draw((-3.0556368653973305,-0.26090424810654206)--(3.558281432597328,-0.2782636137180766), linewidth(0.8)); draw((3.558281432597328,-0.2782636137180766)--(-1.007231723236255,3.4887187239849142), linewidth(0.8)); draw((-1.007231723236255,3.4887187239849142)--(0.25132228359999886,-0.26958393091230937), linewidth(0.8)); draw(circle((-1.3981331930472625,1.2679371918752753), 2.2549222985414907), linewidth(0.8)); draw((0.25132228359999886,-0.26958393091230937)--(1.4816403631598458,1.4351626679090461), linewidth(0.8)); draw((3.212870142514139,3.8339797926548065)--(-1.2857908574057608,-2.3994302224263198), linewidth(0.8)); /* dots and labels */ dot((-1.007231723236255,3.4887187239849142),dotstyle); label("$A$", (-1.084214215315976,3.852353253461407), NE * labelscalefactor); dot((-3.0556368653973305,-0.26090424810654206),dotstyle); label("$B$", (-3.495718857918514,-0.5755507513172563), NE * labelscalefactor); dot((3.558281432597328,-0.2782636137180766),dotstyle); label("$C$", (3.915911230080247,-0.5346777912731456), NE * labelscalefactor); dot((0.25132228359999886,-0.26958393091230937),linewidth(4pt) + dotstyle); label("$D$", (0.30546642618379166,-0.77991555153781), NE * labelscalefactor); dot((1.4816403631598458,1.4351626679090461),linewidth(4pt) + dotstyle); label("$E$", (1.88588754788941,1.3999756508147627), NE * labelscalefactor); label("x", (0.3190907461984953,0.51439484985903), NE * labelscalefactor,qqwuqq); label("x", (0.9458094668748611,0.023919329329701134), NE * labelscalefactor,ffvvqq); label("x", (-2.4466462167863368,0.06479228937381187), NE * labelscalefactor,qqwuqq); label("x", (-1.1659601354041977,2.3400537318293098), NE * labelscalefactor,ffvvqq); label("90 - x", (-0.4711198146543138,2.2719321317557917), NE * labelscalefactor,blue); label("90 - x", (2.226495548257,-0.030577950729113183), NE * labelscalefactor,blue); dot((-0.36867259692921983,-1.1286578414956296),dotstyle); label("$F$", (-0.1713847743308345,-1.4202585922288782), NE * labelscalefactor); label("x", (-0.430246854610203,-0.6164237113613671), NE * labelscalefactor,ffvvqq); dot((3.212870142514139,3.8339797926548065),dotstyle); dot((-1.2857908574057608,-2.3994302224263198),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Note $\angle ADC=2\angle B\implies \angle DAC=\angle A-\angle B\implies DB=DA\implies \angle A = 90^{\circ}$.

Let the angle bisector of $\angle ADC$ meet $AC$ at $E$. Observe $\measuredangle CDE=\measuredangle EDA=\measuredangle DBA$ so $\triangle CDE\sim\triangle CBA$. From this, we obtain that (1) $E$ is thus the midpoint of $\overline{AC}$, implying that $DE\perp AC$ and (2) $DE\parallel BA$. Hence, $AC\perp BA$ as desired.

AlastorMoody wrote: Nice to see that 4 months ago, I wasn't able to solve this problem, now I can do it MENTALLY Let $\ell$ be the angle bisector of $\angle ADC$, Let $X \in \ell$ such that, $AX>BX$, and let $\ell \cap AC=Y$, then, $\angle ABC=\angle ADY$ $=$ $\angle YDC$ $=$ $\angle XDB$ $=$ $\angle DAB$ $\implies$ $BD=CD=AD$ $\implies$ $\angle A=90^{\circ}$ Nice I also solved it MENTALLY it's a very easy angle chase.

What RMO?

how is this rmo

This one was a very simple angle chase. So let the angle bisector of $\angle CDA$ as mentioned in the question intersect side $AC$ of $\triangle ABC$ at $K$. We have $\angle KDC=\angle KDA=\angle DBA=\angle CBA$. This gives $AB \parallel DK$. Now extend $KD$ to an arbitrary point $E$ beyond the side $BC$, we should have $\angle KDC=\angle EDB=angle DAB$. Thus $\triangle DAB$ is isosceles, which means $DA=DB=CD$. This completes our proof.

easy problem.

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