Problem:
Rectangle ABCD is given with AB=63 and BC=448. Points E and F lie on AD and BC respectively, such that AE=CF=84. The inscribed circle of triangle BEF is tangent to EF at point P, and the inscribed circle of triangle DEF is tangent to EF at point Q. Find PQ.
Solution:
Let G and H be the points where the inscribed circle of triangle BEF is tangent to BE and BF, respectively. Let x=EP,y=BG, and z=FH. Then by equal tangents, EG=x,BH=y, and FP=z. Note that y+z=BF=BC−CF=364, and x+y=632+842=212(32+42)=105. Also note that △BEF≅△DFE, so FQ=x. Thus PQ=FP−FQ=z−x=(y+z)−(x+y)=364−105=259.
OR
Let x,y, and z be defined as in the first solution, and let O be the foot of the perpendicular from E to BF. Applying the Pythagorean Theorem to triangle EOF yields EF=EO2+FO2=632+(448−2⋅84)2=72(92+402)=7⋅41=287. Thus z+x=EF=287, and x+y=105 and y+z=364, as shown in the first solution. Adding these three equations together and dividing by 2 yields x+y+z=378. Thus x=378−364=14,y=378−287=91, and z=378−105=273. Therefore PQ=z−x=273−14=259.