Problem:
In quadrilateral ABCD,BC=8,CD=12,AD=10, and m∠A=m∠B=60∘. Given that AB=p+q​, where p and q are positive integers, find p+q.
Solution:
Draw lines containing D and C that are perpendicular to AB at E and F, respectively. Then AE=5,DE=53​,BF=4, and CF=43​. Now draw a line containing C that is perpendicular to DE at G. Because EFCG is a rectangle, GE=CF=43​, so DG=DE−GE=3​. Apply the Pythagorean Theorem to △DGC to find that 141​=GC=EF. Then AB=AE+EF+FB=9+141​, and p+q=150​.
OR
Let P be the intersection of AD and BC, and let AD=a,AB=b,BC=c, CD=d,DP=x, and PC=y. Then △ABP is equilateral, and x+a=y+c= b. Apply the Law of Cosines to △DCP to obtain x2+y2−xy=d2, and then substitute to get (b−a)2+(b−c)2−(b−a)(b−c)=d2. Expand and simplify to get
a2+b2+c2=d2+ab+bc+ac
For the given quadrilateral, this yields 102+b2+82=122+10b+8b+80, and then b2−18b−60=0, whose positive solution is 9+141​. Thus p+q=150​.
