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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.

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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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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Matthew Conroy, Edward Jiang, астон вілла олоф мэллбэрг, JonMark Perry, user91500
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  • $\begingroup$ Welcome, what did you try? Did you sketch a picture, did you add some extra lines or points? $\endgroup$ – Mirko Dec 7 '16 at 1:53
  • $\begingroup$ Hi, i sketched like 20 drawings, all the same thankfully, because i wasn't lost at all, but I spent the whole afternoon thinking about this, and i couldn't arrive at the conclusion, I tried assuming it was true and work backwards but couldn't find a true statement. But well, i am not very good at geometry so that is why i am practicing, and well i wanted to know about how to prove this, because it may yield a step that i was missing in my current knowledge. $\endgroup$ – Enzo Giannotta Dec 7 '16 at 2:08
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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 $\angle CAE = \angle ECA$.

Then

$\angle BAC = \angle BAE + \angle CAE = \angle EBA + \angle ECA = \angle CBA + \angle BCA$.

But also

$\angle BAC + \angle CBA + \angle BCA = 180°$ 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.

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    $\begingroup$ thank you for the answer, i couldnt solve it because i forgot the property that two tangen lines of a circle intersect yielding an isosceles triangle. $\endgroup$ – Enzo Giannotta Dec 7 '16 at 2:18

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