Question

show that a.b.c.d are in proportion if ma2+nb2 : mc2 + nd2 : : ma2- nb2 : mc2 -nd2​

Answer 1

Answer:

hope this is the correct answer

Answer 2

Given:

ma2+nb2 : mc2 + nd2 : : ma2- nb2 : mc2 -nd2

To find:

show that a.b.c.d are in proportion

Solution:

ma²+nb² : mc² + nd² : : ma²- nb² : mc² -nd²

This means that they are in proportion, so the product of extremes should be equal to product of means

So, they can be written as

(ma² + nb²)/(mc² + nd²) =  (ma² - nb²)/ ( mc² - nd²)

Let us assume that a,b,c,d are in proportion, so

a/b = c/d

Squaring on both sides

a²/b² = c²/d²

Multiplying both sides by m/n

ma²/nb² = mc²/ nd²

Applying the rule of adding the numerator and denominator and dividing by subtraction of numerator and denominator on both sides,

(ma² + nb²)/ (ma² - nb²) = (mc² + nd²) / ( mc² - nd²)

Interchanging fractions,

(ma² + nb²)/(mc² + nd²) =  (ma² - nb²)/ ( mc² - nd²)

So, we have arrived at this equation and it is given that this is true so, it can be concluded that our assumption was true.

Hence, a,b,c,d are in proportion.

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