Since the side BC is parallel to AD, that means that the length of the perpendicular from B to AD is equal to the perpendicular from D to BC.
We'll create two right angle triangles ABE and DFC, where BE = DF.
Let the leg AE = x. We'll write Pythagorean theorem in triangle ABE, where AB = 17 is hypotenuse.
BE^2 = 17^2 - x^2
Let the leg FC = 44-16-x = 28-x
We'll write Pythagorean theorem in triangle DFC, where DC = 25 is hypotenuse.
CF^2 = 25^2 - (28-x)^2
But CF = BE=>17^2 - x^2 = 25^2 - (28-x)^2
We'll move the terms in x to the left side:
(28-x)^2 - x^2 = 25^2 - 17^2
We'll expand the binomial:
28^2 - 56x + x^2 - x^2 = 25^2 - 17^2
The difference of two squares form the right side, returns the product:
28^2 - 56x = (25-17)(25+17)
28^2 - 56x = (8)(42)
56x = 784 - 336
56x = 448
x = 8
Now, we'll calculate BE^2 = (17+x)(17-x)
BE^2 = 25*9 = 225
BE = 15
The requested length of the perpendicular drawn from B to AD, is BE = 15.