Verifying a line is tangent to a circle circumscribing a triangle

Question

$\triangle\mathit{ABC}$ is an isosceles triangle with legs $\overline{\mathit{AB}}$ and $\overline{\mathit{BC}}$. $\omega$ is the circle circumscribing the triangle, and $D$ is the intersection of the tangent lines to the circle at $A$ and at $B$. $I$ is the intersection of $\overline{\mathit{CD}}$ and $\omega$. Show that $\overline{\mathit{BD}}$ is tangent to the circle circumscribing $\triangle\mathit{AID}$.

Explanation

Since $\overline{\mathit{DB}}$ is tangent to $\omega$, by the Alternate Segment Theorem, $\angle\mathit{ABD} = \angle\mathit{ACB}$. Since $\angle\mathit{ACB}$ and $\angle\mathit{BAC}$ are the base angles of isosceles triangle $\triangle\mathit{ABC}$, they are equal to each other. So, \begin{equation*} \angle\mathit{ABD} = \angle\mathit{ACB} = \angle\mathit{BAC}. \end{equation*} According to the Alternate Interior Angle Theorem, $\overline{\mathit{DB}}$ and $\overline{\mathit{AC}}$ are parallel. Similarly, since $\overline{\mathit{DA}}$ is tangent to $\omega$, \begin{equation*} \angle\mathit{DAI} = \angle\mathit{ACI} = \angle\mathit{ACD}. \end{equation*} According to the Alternate Interior Angle Theorem, \begin{equation*} \angle\mathit{ACD} = \angle\mathit{BDC} = \angle\mathit{BDI}. \end{equation*} So, $\angle\mathit{DAI} = \angle\mathit{BDI}$.

What theorem can be implemented to conclude that $\overline{\mathit{BD}}$ is tangent to the circle circumscribing $\triangle\mathit{AID}$?

Answer 1

You can use the same theorem that relates inscribed angles and tangents because it's an if and only if statement. Otherwise, you can just say "let $B^\prime$ be such that $DB^\prime$ is tangent to the circle circumscribing $\triangle AID$ and $\overline{DB}=\overline{DB^\prime}$". Then by your theorem $\angle DAI = \angle B^\prime DI$. That means $B^\prime = B$ and we are done.