Question

if (3a+8b) : (3c+8d)=(3a-8b) : (3c-8d) than show that a,b,c,d are in proportion

Answer 1

3a+8b/3c+8d=3a-8b/3c-8d 
(applying alternendo)
=> 3a+8b/3a-8b = 3c+8d/3c-8d
(applying componendo dividendo)(a+b/a-b) = (c+d/c-d)
=>(3a+8b) + (3a-8b) /(3a+8b) -(3a-8b) = (3c+8d) + (3c-8d)/(3c+8d) -(3c-8d)
=> 3a+8b+3a-8b/3a+8b-3a+8b = 3c+8d+3c-8d/3c+8d-3c+8d
=> 6a/16b=6c/16d
=> a/b=c/d
hence,proved!

Answer 2

$\frac{3a+8b}{3c+8d} = \frac{3a-8b}{3c-8d} \\ \\ Cross\ Multiply\ to\ get: \ \ \ (3a+8b)(3c-8d) = (3a-8b)(3c+8d) \\ \\ 9ac-24ad+24bc-64bd = 9ac+24ad-24bc-64bd \\ \\ 48bc = 48ad \\ \\ bc = ad \\ \\ \frac{c}{d} = \frac{a}{b}\ \ \ or, \frac{a}{b} = \frac{c}{d} \\ \\ a: b = c: d$