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There are two circles. There are also two common "outside tangents" (they are marked red in the diagram) and one "inside tangent" which is marked blue. (I am not a native English speaker, I don't know how to call the outside and the inside tangents correctly) The inside tangent intersects the circles in points A and B and intersects two outside tangents in points A1 and B1. How can I prove that the line segment AA1 equals BB1?

Direct&Inverse Tangents

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  • $\begingroup$ The inside tangent intersects the circles in points A1 and B1 Your diagram does not agree with this description ... $\endgroup$ – Antoine Apr 17 '17 at 20:22
  • $\begingroup$ This has been fixed. $\endgroup$ – student28 Apr 17 '17 at 20:24
  • $\begingroup$ What else do you know about the two circles that you didn't mention? How do you know that $B_1B=5.67$ and $AA_1=5.67$? $\endgroup$ – Mercy King Apr 17 '17 at 21:15
  • $\begingroup$ Nothing else. The numbers are just to show that the problem is correct. (measured in GeoGebra). I guess we must find some congruent triangles here. @MercyKing $\endgroup$ – student28 Apr 17 '17 at 21:16
  • $\begingroup$ It is equivalent to show that $AB_1=BA_1$. $\endgroup$ – Jean Marie Apr 17 '17 at 21:22
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Let $E$ be the point of tangency of the lower tangent and the circle on the left. Then $AA_1 = AE$.

Let $F$ be the point of tangency of the lower tangent and the circle on the right.

$AA_1 = EF - A_1B$. $\ BB_1 = E_1F_1 (upper circle) - AB_1$.

Now $AA_1 = A_1B_1 - AB_1$; $BB_1 = A_1B_1 - A_1B$.

From these equations we get that $AB_1 = A_1B$.

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  • $\begingroup$ [+1] I have tried to improve a little your presentation. In particular, use underscores between a letter and its index, example A underscore 1 makes $A_1$. Besides, something is missing in your argumentation: you should mention the symmetry of the figure with respect to axis $A_2C$. $\endgroup$ – Jean Marie Apr 18 '17 at 10:03
  • $\begingroup$ @student28 How is $AA_1 = EF - A_1B$?. $\endgroup$ – Narasimham Apr 18 '17 at 12:51

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