Question

If a, b, c, d are in proportion, then prove that a²+ab+b²/a²-ab+b²=c²+cd+d²/c²-cd+d²

Answer 1

Step-by-step explanation:

In the question,

We have been given that,

a, b, c, d are in proportion.

To Prove :

$\frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}=\frac{c^{2}+cd+d^{2}}{c^{2}-cd+d^{2}}$

Proof:

From the proportion we can say that,

$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$

So,

$b^{2}=ac.......(1)\\c^{2}=bd..........(2)\\d^{2}=ac...........(3)\\a^{2}=bd........(4)\\also,\\ab=cd..........(5)$

On putting the equation (1), (2), (3), (4) and (5) in the LHS of the To Prove equation we get,

$a^{2}=bd\\and,\\c^{2}=bd\\So,\\a^{2}=c^{2}\\Similarly,\\b^{2}=ac\\and,\\d^{2}=ac\\So,\\b^{2}=d^{2}$

Therefore, on putting this in LHS we get,

$\frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}=\frac{c^{2}+cd+d^{2}}{c^{2}-cd+d^{2}}$

Hence, Proved.