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Home » If a, b, c and d are in proportion, prove that: (vii)     (viii)    
If a, b, c and d are in proportion, prove that: (vii)

    \[\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}=\frac{ab+ad-bc}{bc+cd-ad}\]

(viii)

    \[abcd\left[ \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}+\frac{1}{{{d}^{2}}} \right]={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}\]

Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
If a, b, c and d are in proportion, prove that: (vii)

    \[\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}=\frac{ab+ad-bc}{bc+cd-ad}\]

(viii)

    \[abcd\left[ \frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}+\frac{1}{{{d}^{2}}} \right]={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}\]

It is given that

a, b, c, d are in proportion

Consider a/b = c/d = k

a = b, c = dk

Therefore, LHS = RHS.

So we get

= d2 (1 + k2) + b2 (1 + k2)

= (1 + k2) (b2 + d2)

RHS = a2 + b2 + c2 + d2

We can write it as

    \[=\text{ }{{b}^{2}}{{k}^{2}}~+\text{ }{{b}^{2}}~+\text{ }{{d}^{2}}{{k}^{2}}~+\text{ }{{d}^{2}}\]

Taking out the common terms

    \[\begin{array}{*{35}{l}} =\text{ }{{b}^{2}}~({{k}^{2}}~+\text{ }1)\text{ }+\text{ }{{d}^{2}}~({{k}^{2}}~+\text{ }1) \\ =\text{ }({{k}^{2}}~+\text{ }1)\text{ }({{b}^{2}}~+\text{ }{{d}^{2}}) \\ \end{array}\]

Therefore, LHS = RHS.

i

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