Maharashtra State BoardSSC (English Medium) 9th Standard
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If Y + Z a = Z + X B = X + Y C Then Show that X B + C − a = Y C + a − B = Z a + B − C . - Algebra

Sum

 If    `( y + z)/a = ( z + x )/b = ( x + y )/ c`  then show that `x/[ b + c - a ] = y/[ c + a - b ] = z / ( a + b - c)`

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Solution

`( y + z)/a = ( z + x )/b = ( x + y )/ c` 
By invertendo,
`a/( y + z)= b /( z + x )= c/( x + y )` 

`a/( y + z)= b /( z + x )= c/( x + y ) = ( b + c - a)/( z + x + x + y - y - z)`                                         .......( Theorem of equal ratios)

⇒ `a/( y + z)= b /( z + x )= c/( x + y ) = ( b + c - a)/[2x]`    ......(1)

Now,

`a/( y + z)= b /( z + x )= c/( x + y ) = [ c + a - b]/[ x + y +y +z - z -x ]`                                              .......( Theorem of equal ratios)

⇒ `a/( y + z)= b /( z + x )= c/( x + y ) = ( c + b - a)/[2y]`    ......(2)
Also,
`a/( y + z)= b /( z + x )= c/( x + y ) = [ a + b - c]/[ y + z + z + x - x - y ]`                                              .......( Theorem of equal ratios)
⇒ `a/( y + z)= b /( z + x )= c/( x + y ) = ( a + b - c)/[2z]`    ......(3)
From (1), (2) and (3), we have
`[ b + c - a]/(2x) = [ c + a - b]/( 2y) = ( a + b - c)/( 2z)`
⇒ `[ b + c -a ]/x = [ c + a - b]/y = ( a + b - c )/z`
by invertendo,

⇒ ` x / [ b + c -a ]=  y / [ c + a - b]=  z / ( a + b - c )`

Concept: Theorem on Equal Ratios
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APPEARS IN

Balbharati Mathematics 1 Algebra 9th Standard Maharashtra State Board
Chapter 4 Ratio and Proportion
Practice Set 4.4 | Q 3.4 | Page 73
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