In triangle ABC, AB = 10 and AC = 17. Let D be the foot of the perpendicular from A to BC. If BD:CD = 2:5, then find AD.

Let  BC  =  x

So BD   =  (2/7)x

And CD  =  (5/7)x

So

AD  =  √ (AB^2  - BD^2]   = √   ( 10^2  - [ (2/7)x]^2 )   and

AD  =  √  (AC^2  - CD^2 ) = √ ( 17^2 - [ (5/7)x ]^2 )

Since   AD   =  AD......we have that

√   ( 10^2  - [ (2/7)x]^2 )  =  √ ( 17^2 - [ (5/7)x ]^2 )     square  both sides

 ( 10^2  - [ (2/7)x]^2 )  =   ( 17^2 - [ (5/7)x ]^2 )     simplify

100 -  (4/49)x^2  =  289 - (25/49)x^2     rearrange

(25/49)x^2  - (4/49)x^2  =  289 - 100

(21/49)x^2  =  189   

(3/7)x^2 = 189    multiply both sides by  7/3

x^2  =  (7/3)189

x^2  =  63 * 7    

x^2  =  441    

So....AD  =

√  (100  -  (4/49)*441 )  =

√ ( 100 -  (4*9) )  =

√ (100 - 36)  =

√64   =

8