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Question: Let o1, o2 be circles inside an angle tangent to one of its sides in points A, B and to the other in points C, D. Prove that if o1, o2 are externally tangent, then ABCD has an inscribed circle.

What I have so far:

1.|AB| = |CD| by the strongest theorem of geometry

  1. A circle can be inscribed in a quadrilateral if and only if the addition of its opposite sides are equal ie. |AB|+|CD|=|BC|+|AD| (depends on how you label the vertices).

So we want to prove that 2|AB|=|BC|+|AD| and it must have to do with the circles being externally tangent.

I do not know how to proceed. Any help is much appreciated.

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  • $\begingroup$ I'm not sure you have chosen the best result from which to start. There is some symmetry in the situation which may help. $\endgroup$ – Mark Bennet Dec 12 '18 at 8:50
  • $\begingroup$ Please add a diagram. $\endgroup$ – Anubhab Ghosal Dec 12 '18 at 11:10
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Diagram

From the diagram, $AB=CD=(r_1+r_2)\cos\alpha$.
$AC=2r_1\cos\alpha$ and $BD=2r_2\cos\alpha$. Therefore, $AB+CD=AC+BD$, and $ABDC$ is a tangential quadrilateral.

$\blacksquare$

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enter image description here

We can go a little further and show that the common tangent point of the two circles is the center of the inscribed circle of $ABCD$. In deed, by symmetry, we have $\stackrel{\frown}{BE} = \stackrel{\frown}{CE}$. Since $AB$ is tangent to $(O_2)$, it follows that $$\angle ABE = \frac12\stackrel{\frown}{BE} = \frac12\stackrel{\frown}{CE} = \angle EBC.$$ So $E$ lies on the angle bisector of $\angle ABC$. Similarly for all the other angles.

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