2018 AIME I Problems/Problem 15

Suppose our four sides lengths cut out arc lengths of $2a$ , $2b$ , $2c$ , and $2d$ , where $a+b+c+d=180^\circ$ . Then, we only have to consider which arc is opposite $2a$ . These are our three cases, so $\varphi_A=a+c$ $\varphi_B=a+b$ $\varphi_C=a+d$ Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$ , $\overarc{BC}=2b$ , $\overarc{CD}=2c$ , and $\overarc{DA}=2d$ .