This one isn't too bad, tertre !!!
Note that a multiple of a 3-4-5 right triangle is a 15-20-25 right triangle....
Let B = (0,0)
A= (0,15)
C= (20, 0)
The area of ABC is (1/2)BC * BA = (1/2)(20)(15) = 150 which is what we need!!!
Let the circle with the diameter of BC have the equation (x - 10)^2 + y^2 = 100 (1)
And let the slope of the line containing AD = -15/20 = -3/4
And the equation of this line is
y =( -3/4) x + 15 or
y = (15 - 3/4)x
Square both sides of this
y^2 = ( 15 - (3/4)x)^2 (1)
Our objective is to find the x coordinate of the second intersection of AC and this circle
Sub (2) into (1) and we have
(x - 10)^2 + (15 - (3/4)x)^2 = 100 simplify
x^2 -20x + 100 + 225 - (90/4) x + (9/16) x^2 = 100
x^2 -20x + 225 - (90/4) x + (9/16)x^2 = 0 multiply through by 16
16x^2 - 320x + 3600 - 360x + 9x^2 = 0
25x^2 - 680x + 3600 = 0
Believe it or not....we can factor this as
(5x - 36 (5x - 100) = 0
Set each factor to 0 ans solve for x and we have that
x =36/5 and x = 20
We already know that the second value is an x intersection of the line and the circle
We are interested in the first.... and this is the x coordinate of D
The y coordinate is
y = ( - 3/4) (36/ 5) + 15
y = -108/20 + 15
y = -27/ 5 + 75/ 5
y = 48/ 5
So.....D = ( 36/5, 48/5)
So BD is easy to find as
sqrt [ (36/5)^2 + (48/ 5)^2 ] =
sqrt [ 36^2 + 48^2] / 25 =
sqrt [ 3600 ] / 5 =
60 / 5 =
12 units
Note something interesting....that triangle BDC is a 12-16-20 right triangle which is another multiple of a 3-4-5 right triangle !!
Here's a pic of all of this :