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प्रश्न
If `[ax + by]/( x + y) = ( bx + az )/(x + z) = (ay + bz)/[y + z]` and `x + y + z ≠ 0 ` then show that `[a + b]/2`.
उत्तर
Let `[ax + by]/( x + y) = ( bx + az )/( x + z )=( ay + bz )/[ y + z ] ` = k
k = `[( ax + by ) + ( bx + az ) + ( ay + bz )]/[( x +y ) + ( x + z ) + ( y + z )]` ...(By Theorem of equal ratios)
= `[ax + ay + az+bx+by+bz]/ [ 2x + 2y + 2z ]`
= `(a ( x + y + z)+b(x+y+z))/[2( x + y + z )]`
= `(( x + y + z )(a+b))/(2(x + y + z ))`
= `(a + b)/2` ...[x + y + z ≠ 0]
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