If x+y/ax+by = y+z/ay+bz = z+x/az+bx, prove that each of these ratios is equal to 2/a+b, where x+y+z is not = 0
Answers
Answer:
\textbf{Concept used:}Concept used:
\text{If}\;\frac{a}{b}=\frac{c}{d}=\frac{e}{f}If
b
a
=
d
c
=
f
e
\text{then}then
\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\frac{a+c+e}{b+d+f}
b
a
=
d
c
=
f
e
=
b+d+f
a+c+e
\textbf{Given:}Given:
\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}
ax+by
x+y
=
ay+bz
y+z
=
az+bx
z+x
\text{Then, using the given concept}Then, using the given concept
\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{x+y+y+z+z+x}{ax+by+ay+bz+az+bx}
ax+by
x+y
=
ay+bz
y+z
=
az+bx
z+x
=
ax+by+ay+bz+az+bx
x+y+y+z+z+x
\implies\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{2x+2y+2z}{a(x+y+z)+b(x+y+z)}⟹
ax+by
x+y
=
ay+bz
y+z
=
az+bx
z+x
=
a(x+y+z)+b(x+y+z)
2x+2y+2z
\implies\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{2(x+y+z)}{(x+y+z)(a+b)}⟹
ax+by
x+y
=
ay+bz
y+z
=
az+bx
z+x
=
(x+y+z)(a+b)
2(x+y+z)
\implies\bf\frac{x+y}{ax+by}=\frac{y+z}{ay+bz}=\frac{z+x}{az+bx}=\frac{2}{a+b}⟹
ax+by
x+y
=
ay+bz
y+z
=
az+bx
z+x
=
a+b
2