If x/(b + c − a) = y/(c + a − b) = z/(a + b − c), prove that each ratio is equal to (x + y + z)/(a + b + c). Also, show that (b − c)x + (c − a)y + (a − b)z = 0

Question

KnowledgeBoat · Accepted Answer

Given, Let the common value of the given ratios be k. Therefore, x = k(b + c - a), y = k(c + a - b), z = k(a + b - c) Adding x,y and z, we get: ⇒ x + y + z = k[(b + c − a) + (c + a − b) + (a + b − c)] ⇒ x + y + z = k(a + b + c) ⇒ k = Therefore, Given, (b − c)x + (c − a)y + (a − b)z = 0. Substituting value of x, y, z in L.H.S of above equation, we get : ⇒ (b − c)[k(b + c - a)] + (c − a)[ k(c + a - b)] + (a − b)[k(a + b - c)] ⇒ k[(b − c)(b + c - a) + (c − a)(c + a - b) + (a − b)(a + b - c)] ⇒ k[(b2 - c2) - a(b - c) + (c2 - a2) - b(c - a) + (a2 - b2) - c(a - b)] ⇒ k[b2 - c2 - ab + ac + c2 - a2 - bc + ab + a2 - b2 - ca + bc] ⇒ k[b2 - b2 + c2 - c2 + a2 - a2 - ab + ab + ac - ac - bc + bc] ⇒ k(0) ⇒ 0. Hence, proved that and (b − c)x + (c − a)y + (a − b)z = 0.

Mathematics

If x/(b + c − a) = y/(c + a − b) = z/(a + b − c), prove that each ratio is equal to (x + y + z)/(a + b + c). Also, show that (b − c)x + (c − a)y + (a − b)z = 0

Ratio Proportion

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