A problem on a simple result

Relevant wiki: Pitot's Theorem

$\text{A quadrilateral has an inscribed circle (that is, a circle tangent to each of its sides) if and only if the sums of the lengths of the}$ $\text{two pairs of opposite sides are equal. In the above problem, the circle is inscribed in the quadrilateral if and only if the sum}$ $\text{of the lengths of the}$ $\color{#3D99F6}blue~sides$ $\text{is equal to the sum of the lengths of the}$ $\color{#D61F06}red~sides$ . $\text{Therefore, the sum of the lengths of the}$ $\color{#D61F06}red~sides$ $\text{is}$ $\large17+13=$ $\large \boxed{30}$ .

Proof: $\text{Since the tangents to the circle from an external point have the same length, then}$

$AP=AQ, PB=BS, SC=CR$ and $RD=QD$ .

$\text{such that}$

$\color{#3D99F6}AB+CD$ $=AP+PB+CR+RD$

$\color{#3D99F6}AB+CD$ $=AQ+BS+SC+QD$

$\color{#3D99F6}AB+CD$ $=(AQ+QD)+(BS+SC)$

$\color{#3D99F6}AB+CD$ $=$ $\color{#D61F06}AD+BC$

Since $AF, FB, BG, GC, CH, HD, DE, AE$ line segents are tangential, $AF=AE, BF=BG$ , $CG=CH$ , $DH=DE$ . So $AD+BC=AE+ED+BG+GC=AF+BF+CH+DH=AB+CD=\boxed{30}$ .

Pitot thereom says that in a tangential quadrilateral a pair of opposite sides equal the same length as the sum of the other pair of opposite sides. So 17+13=30 thus the sum of the other pair is 30.