Each side of a triangle ABC is 75 units long. Point D is the foot of the perpendicular drawn from A to side BC. Part E is the midpoint of line segment AD. Find BE. Show picture and work.

To find the length of BE, we can use the fact that E is the midpoint of AD.

Since E is the midpoint of AD, we can say that AE = ED.

Now, let's consider triangle ABC. We know that each side of the triangle is 75 units long.

Let's draw a diagram to visualize the triangle.

A
/|\
/ | \
/ | \
/ | \
BE/ | \ AC
/ | \
/______|______\
B AD C

In triangle AED, AE = ED, and let's assume x is the length of AD.

Since E is the midpoint of AD, AE = ED = x/2.

Now, let's consider triangle ABC.

Using the Pythagorean theorem, we can find the length of AC.

AC^2 = AB^2 + BC^2

Since each side of the triangle is 75 units long, we can substitute these values:

AC^2 = 75^2 + 75^2

Simplifying,

AC^2 = 2 * 75^2

AC^2 = 2 * 5625

AC^2 = 11250

Taking the square root of both sides, we get:

AC ≈ √11250

AC ≈ 106.066

Since E is the midpoint of AD, we know that DE = EA = x/2.

So, DC = AC - AD = 106.066 - x.

Let's consider triangle BDC.

Using the Pythagorean theorem, we can find the length of BD.

BD^2 = BC^2 + DC^2

Substituting the values we know:

BD^2 = 75^2 + (106.066 - x)^2

Simplifying,

BD^2 = 5625 + (106.066 - x)^2

Let's consider triangle BDE.

Using the Pythagorean theorem, we can find the length of BE.

BE^2 = BD^2 + DE^2

Substituting the values we know:

BE^2 = 5625 + (106.066 - x)^2 + (x/2)^2

BE^2 = 5625 + 11240.732x - 212.133x + x^2/4

Simplifying,

BE^2 = 16865.732x - 212.133x + x^2/4 + 5625

BE^2 = x^2/4 + 16653.599x + 5625

To find the length of BE, we need to solve for x.

We have a quadratic equation, but it can be simplified since the x^2/4 term is quite small compared to the other terms.

So, we can neglect it for simplicity:

BE^2 ≈ 16653.599x + 5625

To solve for x, we set BE^2 equal to zero:

16653.599x + 5625 = 0

16653.599x = -5625

Divide by 16653.599:

x = -5625/16653.599

x ≈ -0.338

Since x represents the length of AD, which cannot be negative, we disregard this solution.

Therefore, there is no real value for x, which means there is no solution for the length of BE.