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Question
If `(by + cz )/(b^2 + c^2) = (cz + ax)/(c^2 + a^2) = (ax + by)/(a^2 + b^2)` then show that each ratio is equal to `x/a = y/b = z/c`.
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Solution
Each f the given ratio
= `((by + cz))/((b^2 + c^2)) + ((cz + ax))/((c^2 + a^2)) + ((ax + by))/((a^2 + b^2))`
= `(ax + by + cz)/(a^2 + b^2 + c^2)`
Now `(by + cz)/(b^2 + c^2) = (ax + by + cz)/(a^2 + b^2 + c^2)`
⇒ `(a^2 + b^2 + cz)/(b^2 + c^2) = (ax + by + cz)/(by + cz)`
⇒ `(a^2)/(b^2 + c^2) = (zx)/(by + cz)` ...(App. dividendo)
⇒ `(b^2 + c^2)/(a^2) = (by + c^2)/(ax)` ...(App. invertends)
⇒ `(a^2 + b^2 + c^2)/(a^2) = (ax + by + cz)/(ax)` ...(by componendo)
⇒ `x/a = (ax + by + cz)/(a^2 + b^2 + c^2)` ...(similarly)
∴ `x/a = y/b = z/c = (ax + by + cz)/(a^2 + b^2 + c^2)`
Hence proved.
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