Problem:
In △ABC,AB=360,BC=507, and CA=780. Let M be the midpoint of CA, and let D be the point on CA such that BD bisects angle ABC. Let F be the point on BC such that DF⊥BD. Suppose that DF meets BM at E. The ratio DE:EF can be written in the form m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let AB=c,BC=a, and CA=b. Since a>c,F is on BC. Let ℓ be the line passing through A and parallel to DF, and let ℓ meet BD,BE, and BC at D′, E′, and F′ respectively. Since AF′ is parallel to DF,
nm=EFDE=E′F′D′E′
In △ABF′,BD′ is both an altitude and an angle-bisector, so △ABF′ is isosceles with BF′=BA=c. Hence AD′=D′F′, and
Extend BM through M to N so that BM=MN, and draw AN and CN. Quadrilateral ABCN is a parallelogram because diagonals AC and BN bisect each other. Hence AN=BC=a and triangles AE′N and F′E′B are similar. Therefore
1+n2m=E′F′AE′=F′BAN=ca
and
nm=2ca−c=720507−360=24049
so m+n=289.
OR
Let AB=c,BC=a, and CA=b. Let D′ and D′′ be the points where the lines parallel to line DF and containing A and C, respectively, intersect BD, and let E′ and F′, be the points where AD′ meets BM and BC, respectively. Let G be the point on BM so that lines FG and AC are parallel. Note that
ac=DCAD=DD′′DD′=FCFF′=BC−BFBF−BF′=a−BFBF−c
which yields BF=a+c2ac.
Also, MCGF=BCBF, so GF=2b⋅aBF=2b⋅a+c2c=a+cbc.
But ac=DCAD=b−ADAD yields AD=a+cbc.
Therefore AD=GF, which implies that ADFG is a parallelogram, so AG is parallel to DF. Thus G=E′, and then
EFDE=GFDM=ADAM−AD=AD21(AD+CD)−AD=AD21(CD−AD)=21(ADCD−1) so EFDE=21(ca−1)=2ca−c
