Question

Solve 11, ratio problem and give it to me with working.take a pic of your proof.

Answer 1

(y +z)/(Pb + qc) = (z+x)/(pc + qa) = (x+y)/(pa+ qb) = K (let )
then,
(y + z) = K(pb + qc) -------(i)
(z + x) =K (pc + qa)------(ii)
(x + y) = K(pa + qb) -------(iii)

add all eqns

(y + z) + (z + x) + (x + y) = K{P(a + b + c) + q(a + b + c)}
=> 2(x + y + z) = K(a+ b + c)[P + q]
=> 2(x + y + z)/(a + b + c) = K(P + q)----(1)

LHS = 2(x + y + z)/(a + b + c)
from eqn (1)
= K(p + q)

RHS = {(b + c)x + (c + a)y + (a + b)z}/(ab + bc + ca)
= {b( z + x) + c(x + y) + a(y + z)}/(ab + bc + ca)
put eqns (i), (ii), (iii) values ,

= {bK(pc+ qa) + c(pa+qb) + a(pb+qc)}/(ab+bc+ ca)
= K{P(ab+bc+ca) + q(ab+bc+ca)}/(ab+bc+ca)
= K(P+q)(ab+bc+ca)/(ab+bc+ca)
= K(p+q)

Hence, LHS = RHS
proved ///

Answer 2

Let the given equal ratios be = k.

Find LHS and RHS by rearranging the terms in the numerators.

See the proof in the picture enclosed.