∠BAC=90 ,AD⊥BC,BC= 12,
∴ac= 12cosc,ab= 12sinc,bd= 12sin^c,
△ADC is intercepted by BE, and by Menelaus theorem,
AP/PD*DB/BC*CE/EA= 1, P is the midpoint of AD, AE=3,
∴sin^c*( 12cosc-3)/3= 1,
∴( 1-cos^c)(4cosc- 1)= 1,
4cosc- 1-4(cosc)^3+cos^c= 1,
4(cosC)^2-cos^C-4cosC+2=0,①
Let x = cosc ∈ (- 1, 1) and f (x) = 4x 3-x 2-4x+2, then
f'(x)= 12x^2-2x-4= 12(x+ 1/2)(x-2/3),
- 1/2 & lt; X & lt2/3 ft (x)
The minimum value of f (x) = f (2/3) = 32/27-4/9-8/3+2 = (32-12-72+54)/27 = 2/27 > 0,
F (- 1) = 1, F (1) = 1, ∴ F (x) has no zero, that is, equation ① has no solution. There is no solution to this problem.