Math, asked by Anonymous, 4 months ago

if
 \frac{a}{x + y }  \:  =  \frac{b}{y + z}  =  \frac{c}{z - x}
the show that b = a + c ​

Answers

Answered by VishnuPriya2801
56

Answer:-

Let,

 \\   \sf \:  \frac{a}{x + y}  =  \frac{b}{y + z}  =  \frac{c}{z - x}  = k \\

Hence;

 \implies \sf \:  \frac{a}{x + y}  = k \\  \\  \implies \sf \:a = k(x + y) \:  \:  \:  - - equation \: (1). \\  \\

Similarly;

 \implies \sf \:  \frac{b}{y + z}  = k \\  \\  \implies \sf \: b = k(y + z) \:  \:  -  -  \: equation \: (2) \\  \\  \\ \implies \sf \: \frac{c}{z - x}  = k \\  \\  \implies \sf \:c = k(z - x) \:  \:  -  -  \: equation \: (3) \\

Now,

We have to prove;

b = a + c

Substitute the values of a,b and c from equations (1) , (2) & (3).

 \implies \sf \: k(y + z) = k(x + y) + k(z - x) \\  \\  \\ \implies \sf \:k(y + z) = k(x + y + z - x) \\  \\  \\ \implies  \boxed{\sf \:k(y + z) = k(y + z)}

Hence, Proved.

Answered by Anonymous
62

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}}}}}


VishnuPriya2801: Great ! :)
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