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

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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.

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  • $\begingroup$ Are you saying that the Alternate Angle Theorem (or Tangent-Chord Theorem) is a criterion for a line or line segment to be tangent to a circle? $\endgroup$ – user74973 May 11 '20 at 19:19
  • $\begingroup$ 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 $\Gamma$. $\endgroup$ – user74973 May 11 '20 at 19:19
  • $\begingroup$ 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 $\Gamma$. $\endgroup$ – user74973 May 11 '20 at 19:19
  • $\begingroup$ Is that correct? $\endgroup$ – user74973 May 11 '20 at 19:19
  • $\begingroup$ That's right. You can prove the converse using a similar method as described in my answer. You use the theorem with an alternate line that you know is tangent to the circle, and then you prove that line is the same as the line you wanted to prove was tangent. $\endgroup$ – Adrián Delgado May 11 '20 at 19:26

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