Question

(a+b+c+d):(a+b-c-d)=(a-b+c-d):(a-b-c+d) so prove a:b=c:d

Answer 1

consider this friend

Answer 2

Hence Proved.

Step-by-step explanation:

Given,

$(a+b+c+d):(a+b-c-d)=(a-b+c-d):(a-b-c+d)$ then $a:b=c:d$

∴$\frac{(a+b+c+d)}{(a+b-c-d)} =\frac{(a-b+c-d)}{(a-b-c+d)}$

⇒$(a+b+c+d).(a-b-c+d)=(a+b-c-d).(a-b+c-d)$

⇒$a^{2} -ab-ac+ad+ab-b^{2} -bc+bd+ac-bc-c^{2} +cd+ad-bd-cd+d^{2} =a^{2}-ab+ac-ad+ab-b^{2}+ac-bd-ac+bc-c^{2} +cd-ad+bd-cd+d^{2}$(By multiple fraction)

Calculating above equation,

⇒$ad-bc-bc+ad=-ad+bc+bc-ad$

⇒$2ad-2bc=2bc-2ad$

⇒$4ad=4bc$

⇒$ad=bc$

⇒$\frac{a}{b}=\frac{c}{d}$

∴ $a:b=c:d$

Hence Proved.