In quadrilateral ABCD, BC is parallel to AD. E is the foot of the perpendicular from B to AD. Find BE, if AB = 17cm, BC = 16cm, CD = 25cm and AD =...

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.