2013/14 AHSMC Part 1 Problem 8

Since $\angle DAB + \angle CBA = 180^\circ$, $AD$ is parallel to $BC$. Therefore, $\angle ADC + \angle BCD = 180^\circ$. Hence $\angle PDC + \angle PCD = 90^\circ$. Consequently, $\angle DPC = 90^\circ$. Let $Q$ be the foot of the perpendicular on $CD$ from $P$. Note that triangle $PDA$ is congruent to triangle $PDQ$, and triangle $PCB$ is congruent to triangle $PCQ$. Hence $DQ = DA = 8$, $CQ = CB = 18$ and $PA = PQ = PB$. Since $\angle DPC = 90^\circ$, triangle $DPQ$ is similar to triangle $PCQ$. Hence $\dfrac{DQ}{QP} = \dfrac{PQ}{QC}$. Therefore, $PQ^2 = 8 \times 18 = 144$ and $PQ = 12$. The perimeter of $ABCD$ is therefore $AP + PB + BC + CQ + QD + DA = 12 + 12 + 18 + 18 + 8 + 8 = 76$. The answer is (D).