Question

I need explaination also​

Answer 1

$\Large\text{\underline{\underline{Explanation.}}}$

$\large\text{\underline{\underline{Step 1.}}}$

We have -

$\cdots\longrightarrow\boxed{a(y+z)=b(z+x)=c(x+y)=k.}$

We get three equations.

$\cdots\longrightarrow\begin{cases} & y+z=\dfrac{k}{a}.\ \cdots[1]\\\\ & z+x=\dfrac{k}{b}.\ \cdots[2] \\\\ & x+y=\dfrac{k}{c}.\ \cdots[3] \end{cases}$

$\large\text{\underline{\underline{Step 2.}}}$

By $[3]-[2]$, -

$\text{ \cdots\longrightarrow\boxed{y-z=\dfrac{(b-c)ak}{abc}.} }$

By $[1]-[3]$, -

$\text{ \cdots\longrightarrow\boxed{z-x=\dfrac{(c-a)bk}{abc}.} }$

By $[2]-[1]$, -

$\text{ \cdots\longrightarrow\boxed{x-y=\dfrac{(a-b)ck}{abc}.} }$

$\large\text{\underline{\underline{Step 3.}}}$

Then, -

$\text{ \cdots\longrightarrow\boxed{\dfrac{k}{abc}=\dfrac{y-z}{a(b-c)}.} }$

$\text{ \cdots\longrightarrow\boxed{\dfrac{k}{abc}=\dfrac{z-x}{b(c-a)}.} }$

$\text{ \cdots\longrightarrow\boxed{\dfrac{k}{abc}=\dfrac{x-y}{c(a-b)}.} }$

So, -

$\cdots\longrightarrow\dfrac{k}{abc}=\dfrac{x-y}{c(a-b)}=\dfrac{z-x}{b(c-a)}=\dfrac{x-y}{c(a-b)}.$

$\Large\text{\underline{\underline{Final answer}}}$

Hence we proved that, -

$\text{ \cdots\longrightarrow\boxed{\dfrac{x-y}{c(a-b)}=\dfrac{z-x}{b(c-a)}=\dfrac{x-y}{c(a-b)}.} }$