Question

जर `[by + cz ]/[b^2 + c^2] = [cz + ax]/[c^2 + a^2] = [ax + by]/[a^2 + b^2]` तर `x/a= y/b = z/c` हे सिद्ध करा.

Answer 1

`[by + cz]/[b^2 + c^2] = [cz + ax]/[c^2 + a^2] = [ax + by]/[a^2 + b^2]`

समान गुणोत्तरांच्या सिद्धांतावरून,

∴ `[by + cz ]/[b^2 + c^2] = [cz + ax]/[ c^2 + a^2] = [ax + by ]/[a^2 + b^2] = [(by + cz) +(cz + ax) + (ax + by)]/[(b^2 + c^2) + (c^2 + a^2) + (a^2 + b^2)] `

∴ `[2ax + 2by + 2cz ]/[2a^2 + 2b^2 + 2c^2] `

∴ `[2(ax + by + cz) ]/[2(a^2 + b^2 + c^2)]`

∴ `[(ax + by + cz) ]/[(a^2 + b^2 + c^2)] `

याचप्रमाणे आपण सिद्ध करू शकतो, की

∴ `[(by + cz)- (ax + by + cz)]/[(b^2 + c^2) - (a^2 + b^2 + c^2)] = [(cz + ax) - (ax + by + cz)]/[(c^2 + a^2) - (a^2 + b^2 + c^2)]  = [(ax + by) - (ax + by + cz)] /[(a^2 + b^2) - (a^2 + b^2 + c^2)]`

∴ `(-ax)/(-a^2) = (-by)/(-b^2) = (-cz)/(-c^2)`

∴ `x/a = y/b = z/c`