lengths

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 :

Since we know triangle ABC is a 15-20-25 right-angle triangle, then we can easily  find the x coordinate of the second intersection of AC  and this circle as follows:
(150*2) / 25^2 =12/25 x 15 =7.2
(150*2) / 25^2 =12/25 x 20 =9.6, then D =7.2 and 9.6 and:
BD^2 = 7.2^2 + 9.6^2 
BD^2 =144
BD =12