Question

please help standard 9th SSC board ratio and proportion​

Answer 1

Answer:

This is your answer .......

Answer 2

Given:-

• $\sf \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c}$

To show:-

• $\sf \: x(b - c) + y(c - a) + z(a - b) = 0$

Answer:-

Let,

$\sf \dfrac{x}{b + c - a} = \dfrac{y}{c + a - b} = \dfrac{z}{a + b - c} = k$

$\sf \longrightarrow x = k(b + c - a) \quad \quad \dots (1)$

$\sf \longrightarrow y = k(c + a- b) \quad \quad \dots (2)$

$\sf \longrightarrow z = k(a + b- c) \quad \quad \dots (3)$

Required proof:

Considering LHS,

$\sf x(b - c) + y(c - a) + z(a - b)$

Substituting the values of x, y and z from (1), (2) and (3) respectively,

$\sf = k(b + c - a)(b - c) + k(c + a - b)(c - a) + k(a + b - c)(a - b)$

$\sf = k \{(b + c - a)(b - c) + (c + a - b)(c - a) + (a + b - c)(a - b) \}$

$\sf = k \{ b^2 -bc +bc -c^2-ab+ac+ {c}^{2}-ac+ac-a^2-bc + ab + {a}^{2} -ab +ab -b^2 -ac +bc \}$

$\sf = k \times 0$

$\sf = 0$

Considering RHS,

$\sf = 0$

So L.H.S. = R.H.S.

Hence proved!