數學題 高人指點

Let OH and OK be the perpendicular lines from centre O to AB and AC respectively.Also, let DM and EN be the perpendicular lines from D and E to BC.
For triangle HOD and triangle DBM:
OH = DM (radius of circle)
angle OHD=angle DMB (right angles)
angle HDO=angle DBM (corres. angle DE//BC)
therefore, the 2 triangle are congruent.
therefore DH=BM and OD=BD........................(1)
Similarly, triangle KOE is congruent to triangle ENC,
therefore, KE=NC and OE=EC.........................(2)
Also, AH=AK (tangents to circle)......................(3)
AB + AC = 5 + 7= 12
that is AH + DH + BD + AK + KE + EC = 12
Using the results of (1), (2) and (3), we get
AH + DH + OD + AK + KE + OE = 12, but OD + OE = DE = 4, therefore,
AH + DH + AK + KE = 8 , that is AD + AE = 8.
Now referring to the 2 similar triangles ADE and ABC,
AD + AE = 8, while AB + AC = 5 + 7=12, therefore, the ratio is 12/8=1.5
that means BC/DE also=1.5, therefore, BC=1.5 x DE= 1.5 x 4=6.