Componendo et Dividendo |
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The following theorems on proportion are
very useful. The names of the theorems in Latin are included for interest only.
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Proofs of Componendo et Dividendo (1) k-method Let Then a = bk, c = dk (2) Combine the proofs of
Componendo and Dividendo theorems Divide
(1) by (2) and cancel the “b” and “d” in the denominators of (1) and (2), (3) Divide and multiply Given: |
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Converse of Componendo et Dividendo The point is : Can we go
backward? Can we prove: |
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Proofs of the converse (1) Direct expansion (a
+ b)(c – d) = (a – b)(c + d) ac
– ad + bc – bd = ac + ad – bc – bd \ 2bc = 2ad ad = bc (2) Apply Componendo et Dividendo itself This
is quite interesting. You use the theorem to prove the converse of the
theorem! |
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Exercise Apply
Componendo et Dividendo to the following proportional and solve the equation: Highlight the
box below for answer:
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