Estonian Math Competitions

Let the line $AD$ intersect the circumcircle of the triangle $ABC$ a second time at point $D^{\prime }$; by the assumptions, $\angle CD{D}^{\prime }=\angle D^{\prime }DB=\angle CAB$ (see figure below).



We show that $AD=D{D}^{\prime }$. As the quadrilateral $AB{D}^{\prime }C$ is cyclic, $\angle ABC=\angle A{D}^{\prime }C=\angle D{D}^{\prime }C$ and $\angle BCA=\angle B{D}^{\prime }A=\angle B{D}^{\prime }D$. Thus the triangles $ABC$, $D{D}^{\prime }C$ and $DB{D}^{\prime }$ are similar. Hence $\frac{|D{D}^{\prime }|}{|DB|}=\frac{|DC|}{|D{D}^{\prime }|}$, implying $|D{D}^{\prime }{|}^2=|BD|\cdot |CD|$.

By similarity of the triangles $ABC$ and $D{D}^{\prime }C$, we obtain $\angle BCA=\angle D^{\prime }CD$, which implies $\angle DCA=\angle D^{\prime }CB=\angle D^{\prime }AB=\angle DAB$. By similarity of the triangles $ABC$ and $DB{D}^{\prime }$ we analogously get $\angle ABD=\angle CB{D}^{\prime }=\angle CA{D}^{\prime }=\angle CAD$. Hence the triangles $ABD$ and $CAD$ are similar. Consequently, $\frac{AD}{CD}=\frac{BD}{AD}$, which implies $A{D}^2=BD\cdot CD$.

Altogether, we have proven $A{D}^2=D{D}^{\prime 2}$, which implies the desired result.

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Alternative solution.

Let the line $AD$ intersect the circumcircle of the triangle $ABC$ a second time at point $D^{\prime }$; by the assumptions, $\angle CD{D}^{\prime }=\angle D^{\prime }DB=\angle CAB$. We show that $AD=D{D}^{\prime }$.

Let $O$ be the circumcenter of the triangle $ABC$. Let the line $AD$ intersect the circumcircle of the triangle $BCD$ a second time at point $A^{\prime }$ (see figure below).



By the equality $\angle CD{A}^{\prime }=\angle A^{\prime }DB$, the arcs $B{A}^{\prime }$ and $A^{\prime }C$ of the circumcircle of the triangle $BCD$ are equal.

By the conditions of the problem, $\angle CDB=2\angle CAB$. Hence point $O$ lies on the circumcircle of the triangle $BCD$. Since $OC=OB$, the arcs $CO$ and $OB$ of the circumcircle of the triangle $BCD$ are equal.

Consequently, $O{A}^{\prime }$ is a diameter of the circumcircle of the triangle $BCD$. If $O\ne D$ then, by Thales' theorem, $\angle A^{\prime }DO=90^{\circ }$. Thus $OD\perp A{D}^{\prime }$, which implies $AD=D{D}^{\prime }$ since a radius perpendicular to a chord bisects the chord. If $O=D$ (see figure below) then $A{D}^{\prime }$ is a diameter of the circumcircle of the triangle $ABC$, obviously bisected by the center $O$.