If \(x, y, z\) are the distances of the vertices of the points \(A,B\) and \(C\) respectively in \(ΔABC\) from the orthocentre, then simplify the expression:

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\)

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## Comments

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TopNewestWe know that : \(x = 2RcosA \) , \( y = 2RcosB \) and \(z = 2RcosC \)

\(\frac{a}{x} + \frac{b}{y} + \frac{c}{z} \)

= \(\frac{2RsinA}{2RcosA} + \frac{2RsinB}{2RcosB} + \frac{2RsinC}{2RcosC} \)

=\( tanA + tanB + tanC \)

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Nice attempt.. But if the options were the following then what would be the solution???:

\((a)\frac{xyz}{abc}\)

\((b)\frac{(xy+yz+zx)^{3}}{(abc)^{2}}\)

\((c)\frac{abc}{xyz}\)

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i think its option- c. since A+B+C=180. THUS tanA +tanB +tanC =tanAtanBtanC. but now clearly tanA=(a/x) tanB=(b/y) and tanC=(c/z). therefore, (a/x)+(b/y)+(c/z)=(abc/xyz). which is a pretty obvious.

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Thanks.. but could you tell me a good book/website would i find these formulae(those of properties of triangles) with their proof??

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