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In a circumscribed quadrilateral, the two pairs of opposite sides add up to the same total length: $a+c=b+d$. Conversely, any quadrilateral with $a+c=b+d$ must be circumscribed.

Suppose we are given four lengths $a,b,c,d$ with $a+c=b+d$, and moreover write $a=a_1+a_2$. Is it true that there exists a quadrilateral with consecutive side lengths $a,b,c,d$ such that the tangent point of the incircle to the side with length $a$ breaks the side into two segments of length $a_1$ and $a_2$?

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Yes, it is true.

When you have the eight tangent lengths, the angle formulas enable you to find the angles of such a quadrilateral.

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    How do you know that any tangent lengths give rise to such a quadrilateral? – 2017-02-21
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    @pi66: The point is that the angle formulas tell us that the angles of such a quadrilateral are determined only by the eight tangent lengths. We can construct a quadrilateral satisfying (1)(2). (1) The side lengths are $a=a_1+a_2, b,c,d$ with $a+c=b+d$ (it follows that the quadrilateral has an incircle, but note that the angles are not determined yet). (2) The angles are determined by the eight tangent lengths $a_1,a_2,a_2,b-a_2,b-a_2,c-b+a_2,c-b+a_2,d-c+b-a_2$ where $d-c+b-a_2=a_1$. – 2017-02-22