Problem: A circle $\omega$ is inscribed in a quadrilateral $ABCD$.
Let $I$ be the center of $\omega$.
Suppose that \[ (AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2.  \]
Prove that $ABCD$ is an isosceles trapezoid.

\end{enumerate}

Solution: Here's a completely algebraic solution.
WLOG $\omega$ has radius $1$, and let $a$, $b$, $c$, $d$ be the
lengths of the tangents from $A$, $B$, $C$, $D$ to $\omega$.
It is known that
\[ a+b+c+d = abc+bcd+cda+dab \qquad (\star) \]
which can be proved by, say $\tan$-addition formula.
Then, the content of the problem is to show that
\[
  (\sqrt{a^2+1}+\sqrt{d^2+1})^2
  + (\sqrt{b^2+1}+\sqrt{c^2+1})^2
  \le (a+b+c+d)^2
\]
subject to $(\star)$,
with equality only when $a=d=\tfrac1b=\tfrac1c$.

Let $S = ab+bc+cd+da+ac+bd$. Then the inequality is
\[ \sqrt{(a^2+1)(d^2+1)} + \sqrt{(b^2+1)(c^2+1)} \le S - 2. \]
Now, by \textbf{USAMO 2014 Problem 1} and the condition $(\star)$,
we have that $(a^2+1)(b^2+1)(c^2+1)(d^2+1) = (S - abcd - 1)^2$.
So squaring both sides, the inequality becomes
\[ (ad)^2+(bc)^2 + a^2+b^2+c^2+d^2 \le S^2 - 6S + 2abcd + 4. \]
To simplify this, we use the identities
\begin{align*}
  S^2 &= 6abcd + \sum_{\text{sym}} a^2bc
    + \frac14\sum_{\text{sym}} a^2b^2 \\
  (a+b+c+d)^2 &= (abc+bcd+cda+dab)(a+b+c+d) \\
  &= 4abcd + \frac12\sum_{\text{sym}} a^2bc
\end{align*}
So $S^2+2abcd = \frac14\textstyle\sum_{\text{sym}}
  a^2b^2 + 2(a^2+b^2+c^2+d^2) + 4S$
and the inequality we want to prove reduces to
\[ 2S \le (ab)^2+(ac)^2+(bd)^2+(cd)^2 + 4 + a^2 + b^2 + c^2 + d^2. \]
This follows by AM-GM since
\begin{align*}
  (ab)^2 + 1 &\ge 2ab \\
  (ac)^2 + 1 &\ge 2ac \\
  (bd)^2 + 1 &\ge 2bd \\
  (cd)^2 + 1 &\ge 2cd \\
  a^2 + d^2 &\ge 2ad \\
  b^2 + c^2 &\ge 2bc.
\end{align*}
The equality case is when $ab=ac=bd=cd=1$, $a=d$, $b=c$,
as needed to imply an isosceles trapezoid.

\begin{remark*}
Note that a priori one expects an inequality.
Indeed,
\begin{itemize}
  \ii Quadrilaterals with incircles have four degrees of freedom.
  \ii There is one condition imposed.
  \ii Isosceles trapezoid with incircles have two degrees of freedom.
\end{itemize}
\end{remark*}