Question 1140950
{{{BD}}} is the altitude of right triangle {{{ABC}}}. 

If {{{AD=3 }}}and {{{DC=12}}}, what is the length of {{{BD}}}?


{{{drawing( 600, 600, -10, 20, -10, 20,
circle(0,8,.12),locate(0.5,8,A),
circle(0,0,.12),locate(0.5,-0.5,B),
circle(0,12,.12),locate(12,0.5,C),
line(0,8,12,0),line(0,0,3,6),
circle(3,6,.12),locate(3,6,D),locate(2,7.5,3),locate(6,5,12),
locate(0.5,1,1),locate(0.2,1.5,2),locate(0.2,7,1),locate(10,0.7,2),
 graph( 600, 600, -10, 20, -10, 20, 0)) }}}



given:  right triangle {{{ABC}}}=> angle {{{B = 90}}}°  and altitude divides angle {{{B}}} into two angles ,  < {{{1}}} and  < {{{2 }}}

=> <{{{1 + 2 = 90}}}°  ……(1)

{{{BD}}} is perpendicular to{{{ AC }}}( given)

=> <{{{1}}} + < {{{DCB = 90}}}° …….(2)

=> < {{{DCB }}}= < {{{2}}} ………….by (1) & (2)

& < {{{BAD}}} = <{{{1}}}

So, triangle{{{ BCD}}} ~ triangle{{{ ABD}}} ( by AAA similarity theorem)

=> {{{BC/AB = CD/BD = BD/AD}}} ( corresponding sides of similar triangles)

=> {{{BC/AB = 12/BD = BD/3}}}

= {{{BD^2 = 36}}}

=> {{{BD = sqrt(36)}}}

=> {{{BD = 6}}}