Question 476259
{{{drawing(400,360,-5,15,-7,11,
locate(0,6.7,A), locate(0,0,B), locate(8,0,C),
locate(-.6,1.2,x), locate(-.6,4,4), locate(.8,.1,x), locate(1.7,5.5,4),
locate(3.2,0,2x+2),locate(5.4,2.5,2x+2),
circle(0,0,2), circle(8,0,6), circle(0,6,4), triangle(0,0,8,0,0,6) )}}}
<pre>
AB = 4+x
BC = x + (2x+2) = x + 2x+2 = 3x + 2
AC = 4 + (2x+2) = 4 + 2x+2 = 2x + 6

By the Pythagorean theorem:

                 AB² + BC² = AC²

          (4+x)² + (3x+2)² = (2x+6)²

 (4+x)(4+x) + (3x+2)(3x+2) = (2x+6)(2x+6)

  (16+8x+x²) + (9x²+12x+4) = 4x²+24x+36

        16+8x+x²+9x²+12x+4 = 4x²+24x+36

               10x²+20x+20 = 4x²+24x+36

                 6x²-4x-16 = 0

Divide through by 2

                  3x²-2x-8 = 0

That was the first thing you were asked to show.

Factor the left side:

               (3x+4)(x-2) = 0

            3x+4 = 0       x-2 = 0
              3x = -4        x = 2
               x = -4/3

We ignore the negative answer. 
So the value of x is 2.  That was the second 
thing you wanted. 

We calculate the lengths of the sides:
  
AB = 4+x = 4+2 = 6
BC = 3x+2 = 3(2)+2 = 6+2 = 8
AC = 2x+6 = 2(2)+ 6 = 4+6 = 10

{{{drawing(400,360,-5,15,-7,11,
locate(0,6.7,A), locate(0,0,B), locate(8,0,C),
locate(-.6,1.2,2), locate(-.6,4,4), locate(.8,.1,2), locate(1.7,5.5,4),
locate(3.2,0,6),locate(5.4,2.5,6), 
circle(0,0,2), circle(8,0,6), circle(0,6,4), triangle(0,0,8,0,0,6) )}}}

Now we want to find the length of the perpendicular from 
B to AC.  We'll draw that perpendicular AD in and label 
it h (in green):

{{{drawing(400,360,-5,15,-7,11,
locate(0,6.7,A), locate(0,0,B), locate(8,0,C), green(line(0,0,2.88,3.84)),
green(locate(1,2.7,h),locate(3,4.6,D)),

locate(-.6,1.2,2), locate(-.6,4,4), locate(.8,.1,2), locate(1.7,5.5,4),
locate(3.2,0,6),locate(5.4,2.5,6), 
circle(0,0,2), circle(8,0,6), circle(0,6,4), triangle(0,0,8,0,0,6) )}}}

Now the circles just get in the way, so we'll erase them
and just look at the triangle, whose sides are

AB=6, BC=8, AC=10

So we can dispense with the circles and just keep the triangle and
the green altitude AD = h:

{{{drawing(400,360,-5,15,-7,11, green(line(0,0,2.88,3.84)),
locate(0,6.7,A), locate(0,0,B), locate(8,0,C),  locate(-.6,3,6), 
green(locate(1,2.7,h), locate(3,4.6,D)),
locate(3.4,0,8), triangle(0,0,8,0,0,6) )}}}

A perpendicular from the right angle to the hypotenuse of a right 
triangle divides the right triangle into two smaller similar right
triangles, both of which are similar to the whole right triangle.
So

&#4416;ADB&#8764;&#4416;BDC&#8764;&#4416;ABC

Using the fact that &#4416;ADB&#8764;&#4416;ABC and the fact that AC=10

{{{BD/AB=CB/AC}}}

{{{h/6=8/10}}}

Cross-multiply:

{{{10h=48}}}

{{{h = 4.8}}}

So the length of the perpendicular AD is 4.8

Edwin</pre>