Does the quadrilateral have an inscribed circle?

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.

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

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.
$$\blacksquare$$
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.