Question 745364

if {{{AC=(4/3)AB}}} => {{{3AC=4AB}}} => {{{AB=(3/4)AC}}}

and since  {{{BC =AC-AB}}} we can say:

{{{BC =(1/4)AC}}}

therefore  {{{AB=3BC}}}
 
furthermore, {{{CD= BD- BC}}}

substituting: {{{CD = 6BC - BC}}}  =>  {{{CD =5BC}}}

now we have both {{{AB}}} and {{{CD}}} in terms of {{{BC}}}, so:

{{{AB/CD =3BC/5BC =3/5}}}