Question

$if \ \textless \ br /\ \textgreater \ \frac{a}{x + y } \: = \frac{b}{y + z} = \frac{c}{z - x}  \\ \ \textless \ br /\ \textgreater \ then \: show \: that \: b = a + c ​$
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Answer 1

Answer:-

Given:-

$\leadsto$ $\sf \dfrac{a}{x + y} =\: \dfrac{b}{y + z} =\: \dfrac{c}{z - x}$

Prove That :-

$\leadsto$ $\sf b =\: a + c$

Solution:-

$\implies \sf \dfrac{a}{x + y} =\: \dfrac{b}{y + z} =\: \dfrac{c}{z - x} =\: k\: [where\: k \ne 0]$

Then, we get

↦ $\sf \dfrac{a}{x + y} =\: k$

By doing cross multiplication we get,

➦ $\sf\bold{\pink{a =\: k(x + y)}}$

↦ $\sf \dfrac{b}{y + z} =\: k$

By doing cross multiplication we get,

➦ $\sf\bold{\pink{b =\: k(y + z)}}$

And,

↦ $\sf \dfrac{c}{z - x} =\: k$

By doing cross multiplication we get

➦ $\sf\bold{\pink{c =\: k(z - x)}}$

Hence, we get the value of a , b and c :

→a = k(x + y)

→b = k(y + z)

→c = k(z - x)

Then,

$\longmapsto$ $\sf b =\: a + c$

By putting the value of a , b and c we get,

⇒ $\sf k(y + z) =\: k(x + y) + k(z - x)$

⇒ $\sf k(y + z) =\: k(\cancel{x} + y + z \cancel{- x})$

➠ $\sf\boxed{\bold{k(y + z) =\: k(y + z)}}$

$\longrightarrow$ ${\red{\bigstar}} \ {\underline{\green{\textsf{\textbf{HENCE, PROVED}}}}}$