Funky Quadrilaterals

Since $\angle AFP=\angle PDC$ , $\angle APF=\angle DPC$ and $DP=PF=18$ , $\therefore \triangle APF\cong \triangle CPD$ $\therefore AP=PC$ Since $ABCD$ is cyclic, we have $AP\times PC=BP\times PD=8\times18=144$ $\therefore AP=PC=12$ On the other hand, we have $\angle AFD=\angle BDC=\angle BAC=\angle AED$ , this implies that $AFED$ is also cyclic, thus $\angle PBC=\angle DAE=\angle DFE$ $\therefore BC\parallel FE$ $\therefore \triangle PBC \sim\triangle PFE$ $\therefore \frac{PB}{BF}=\frac{PC}{CE}$ $\frac{8}{10}=\frac{12}{CE} \\CE=15$ Finally, by Ptolemy's theorem , we have $\begin{aligned}AF\times DE+AD\times FE&=AE\times FD \\&=(12+12+15)\times36 \\&=\boxed{1404} \end{aligned}$

/-CDB = /-CAB (same sector). AF//DC so /-CDB=/-AFB. In other words, /-CAB=/-AFB. This implies that PA is tangent to the circumcircle of Tr. AFB or PA²= PB x PF = 8 x 18 or PA =12.

Now, AP x PC = DP x PB or 12 x PC =18 x 8 or PC = 12.

In a similar fashion, PC x PE =18² which gives us, 12(12+CE) =18² or CE = 15.

We now observe that PB/BF = 8/10 = PC/CE or BC//FE which implies that /-ACB=/-AEF.

But /-ACB=/-ADF. In other words, /-AEF = /-ADF which, in turn, means that ADEF is concyclic. Hence, the required expression, AF x DE + AD x FE = AE x DF = (12+12+15)(18+8+10) = 1404