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)`
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 )`
