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| Properties of proportions |
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| If a:b=c:d then b:a = d:c |
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| a:b=c:d |
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| Dividing 1 by each of these ratios, |
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| If a:b=c:d, then a:c=b:d. |
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| a:c=b:d |
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| If a:b=c:d, then a+b:b=c+d:d |
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| Adding 1 to both sides, we get |
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| If a:b=c:d then a-b:b=c-d:d |
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| Subtracting 1 from both sides, we get |
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| If a:b=c:d then a+b:a-b::c+d:c-d. |
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| a:b=c:d |
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 a+b:b=c+d:d by componendo |
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| a-b:b=c-d:d by dividendo |
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| Dividing, |
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| 1) If a:b::c:d, show that 2a+3b:2a-3b::2c+3d:2c-3d. |
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| Suggested answer: |
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| a:b::c:d |
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| Multiplying both sides by 2/3, we get |
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| Applying componendo and dividendo, |
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| 2a+3b:2a-3b=2c+3d:2c-3d |
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| Suggested answer: |
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| By alternendo, |
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| Applying Componendo/Dividendo, we get |
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| we get |
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| Dividing both sides by 2x, we get |
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| By componendo and dividendo, we get |
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| Adding (1) and (2), |
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| =2 |
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| 4) If a, b, c and d are in continued proportion then prove that |
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| Suggested answer: |
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| a, b, c, d are in continued proportion. |
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| then a=bk, b=ck, c=dk |
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| Consider, |
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| Consider, |
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| Suggested answer: |
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| then x=ak, y=bk, z=ck |
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| Consider, |
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| =k3/2 |
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| Suggested answer: |
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| By componendo and dividendo, we get |
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| Squaring both sides, |
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| By componendo and dividendo, we get |
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| x(b2+1)=2ab |
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| Suggested answer: |
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| (a2+b2) (m2+n2)=(am+bn)2 given |
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| a2m2+b2m2+a2n2+b2n2=a2m2+b2n2+2ambn |
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| b2m2+a2n2-2ambn=0 |
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| (bm-an)2=0 |
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 bm=an |
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