Помогите с геометрией

1) AB=BC=AC=36/3=12
AD=CD=(40-12)/2=14

1) AB = BC = AC = x
AD = DC = y
P (ABC) = AB + BC + AC = x + x + x = 3x = 36 --->
x = 36\3 = 12
P (ADC) = AD + DC + AC = y + y + x = 2y + x = 2y + 12 = 40
2y = 40 - 12 = 28 --->
y = 28\2 = 14
AB = BC = AC = x = 12
AD = DC = y = 14
2) < BQE = < BEQ ---> QB = BE
BQ = 3,5*QE
P (QBE) = BQ + BE + QE = 6,4 =>
2*BQ + QE = 6,4
2*(3,5*QE) + QE = 6,4
8*QE = 6,4
QE = 6,4\8 = 0,8
BQ = BE = 3,5*QE = 3,5*0,8 = 2,8
3) AB = BC; AC = 8
BE = EC = x ----> AB = BC = 2x
P (ABE) = AB + BE + AE = 2x + x + AE = 3x + AE
P (ACE) = AC + CE + AE = 8 + x + AE
P (ABE) = P (ACE) - 2
=>
3x + AE = (8 + x + AE) - 2
3x - x = 8 - 2
2x = 6 ---> BE = EC = x = 6\2 = 3
AB = BC = 2x = 2*3 = 6