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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  • $\begingroup$ 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

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 $\angle CAE = \angle ECA$.


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

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.

  • 1
    $\begingroup$ 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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