# Verifying a line is tangent to a circle circumscribing a triangle

$$\triangle\mathit{ABC}$$ 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 $$\triangle\mathit{AID}$$?

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.
• From the description in the post, $\overline{\mathit{AD}}$ and $\overline{\mathit{DI}}$ are chords of the circle $\Gamma$ circumscribing $\triangle\mathit{AID}$. We are trying to verify that $\overline{\mathit{BD}}$ is tangent to $\Gamma$. May 11, 2020 at 19:19
• Since the equality $\angle\mathit{DAI} = \angle\mathit{BDI}$ has been verified, we can conclude from the Alternate Segment Theorem that $\overline{\mathit{BD}}$ is tangent to $\Gamma$. May 11, 2020 at 19:19