Does the quadrilateral have an inscribed circle?

Question

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

2. 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.

Answer 1

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.

$\blacksquare$

Answer 2

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.