IRL_ABooklet

Let $T$ be the point on the line $CD$ for which $TA$ is the common tangent to both circles. Then $|TA|=|TB|$, because $TB$ is a tangent to the smaller circle as well. Hence $\angle TBA=\angle TAB$.



Looking at triangle $ABC$ we see that $\angle TBA=\angle BDA+\angle BAD$. Since $\angle TAB=\angle CAB+\angle TAC$ and $\angle TAC=\angle BDA$ by the alternate segment theorem, we get $\angle TAB=\angle CAB+\angle BDA$. Comparing the two expressions for $\angle TBA=\angle TAB$ we obtain $\angle CAB=\angle BAD$, i.e. $AB$ bisects $\angle CAD$.

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Alternative solution.

The points $A$, $I$, and $J$ are collinear since $IA$ and $JA$ are both perpendicular to the common tangent at $A$. Let $E$ be the second intersection point of $AB$ with the circle centre $J$. Join $IB$ and $JE$.



Because $|IA|=|IB|$ we have $\angle ABI=\angle IAB$, and from $|JA|=|JE|$ we get $\angle AEJ=\angle JAE$. This implies $\angle ABI=\angle AEJ$, hence $IB$ is parallel to $JE$. As $IB$ is perpendicular to the tangent $CD$, it follows that $JE$ is perpendicular to the chord $CD$, i.e. $E$ is the midpoint of the arc $CD$ and $\angle CAE=\angle DAE$, as required.