Two circles are tangent externally at A, and a common external tangent touches them at B and C. The line segment BA is extended, meeting the second circle at D. Prove that CD is a diameter.
closed as off-topic by Matthew Conroy, Edward Jiang, астон вілла олоф мэллбэрг, JonMark Perry, user91500 Dec 7 '16 at 6:14
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Let $E$ be the point where $BC$ and the common tangent through $A$ intersect. $|AE|=|BE|$ for the two tangent segments from $E$ to the circle containing $A$ and $B$, thus triangle $AEB$ is isosceles with $\angle BAE = \angle EBA$. Likewise $\angle CAE = \angle ECA$.
$\angle BAC = \angle BAE + \angle CAE = \angle EBA + \angle ECA = \angle CBA + \angle BCA$.
$\angle BAC + \angle CBA + \angle BCA = 180°$ in triangle $ABC$.
Then $\angle BAC = 90°$, and the inscribed $\angle CAD$ which is supplementary to $\angle BAC$ is also $90°$. This identifies the chord $CD$ as a diameter.