umm.. is that a trick question? cos i got it in like 5 lines... well 5 of my lines anyway. ok quadrilateral ABCD, draw the bisectors of angle A and angle D, where they meet can be called E (because E comes after D. how clever)
angle BAE = angle DAE (AE is the bisector of angle BAD. heh bad)
angle CDE = angle ADE (DE is the bisector of angle ADC. still following? yeah? good)
let angle EAD = a and angle EDA = b (i used alpha and beta but they're stupid to type on computer)
so angle BAD = 2a and angle ADE = 2b
the sum of the remaining angles = 360 - 2a - 2b (sum of the angles in a quadrilateral)
= 2(180 - a - b)
but angle AED = 180 - a - b (sum of the angles in a triangle)
therefore angle AED = 1/2(angle B + angle C)
done and done. but it was easier when i wrote it on a piece of paper
