b+c-a/y+z-x=c+a-b/z+x-y=a+b-c/x+y-z, then prove that a/x=b/y=c/z
Answers
Step-by-step explanation:
Let &
Each ratio = sum of antecedents ÷ sum of consequents. Here, each ratio is:
=>
=>
=>
=>
=>
=>
•
•
•
=>
=>
Proved.
Step-by-step explanation:
Step-by-step explanation:
Let a + b + c = Aa+b+c=A & x + y + z = Bx+y+z=B
Each ratio = sum of antecedents ÷ sum of consequents. Here, each ratio is:
=> \small{\frac{(b+c-a)+(c+a-b)+(a+b-c)}{(y+z-x)+((z+x-y)+(x+y-z)} }
(y+z−x)+((z+x−y)+(x+y−z)
(b+c−a)+(c+a−b)+(a+b−c)
=> \frac{a+b+c}{x+y+z}
x+y+z
a+b+c
=> \frac{A}{B}
B
A
And, hence,And,hence,
=> \frac{b+c-a}{y+z-x}=\frac{c+a-b}{z+x-y }=\frac{a+b-c}{x+y-z}=\frac{A}{B}
y+z−x
b+c−a
=
z+x−y
c+a−b
=
x+y−z
a+b−c
=
B
A
=> \frac{a+b+c-2a}{x+y+z-2x}=\frac{a+b+c-2b}{x+y+z-2y} =\frac{a+b+c-2c}{x+y+z-2z}=\frac{A}{B}
x+y+z−2x
a+b+c−2a
=
x+y+z−2y
a+b+c−2b
=
x+y+z−2z
a+b+c−2c
=
B
A
=> \frac{A-2a}{B-2x}=\frac{A-2b}{B-2y}=\frac{A-2c}{B-2z}=\frac{A}{B}
B−2x
A−2a
=
B−2y
A−2b
=
B−2z
A−2c
=
B
A
Then, we\: can\:say,Then,wecansay,
• A - 2a = \frac{A}{B}(B - 2x) → \frac{a}{x} = \frac{A}{B}A−2a=
B
A
(B−2x)→
x
a
=
B
A
• A - 2b =\frac{A}{B} (B - 2y) → \frac{b}{y} = \frac{A}{B}A−2b=
B
A
(B−2y)→
y
b
=
B
A
•A - 2c =\frac{A}{B} (B - 2z) →\frac{c}{z}=\frac{A}{B}A−2c=
B
A
(B−2z)→
z
c
=
B
A
Compare \:all,Compareall,
=> \frac{a}{x} =\frac{b}{y} =\frac{c}{z}=\frac{A}{B}
x
a
=
y
b
=
z
c
=
B
A
=>\frac{a}{x} =\frac{b}{y} =\frac{c}{z}
x
a
=
y
b
=
z
c
Proved.