An Important Property –
a c e
If, ------ = ------- = ------- = k,
b d f a + c + e
prove that each ratio is equal to ------------
b + d + f
sum of antecedent
i.e., each ratio = --------------------
sum of consequent
a c e
Proof – Let, ------- = -------- = -------- = k.
b d f
Then, a = bk, c = dk, and e = fk
a + c + e bk + dk + fk
So, ------------ = ---------------
b + d + f b + d + f
k (b + d + f)
= --------------- = k
(b + d + f)
a + c + e
Hence, each ratio = ------------
b + d + f
x + y y + z z + x
Example.1) If, --------- = --------- = ---------, prove that each of
ax + by ay + bz az + bx
2
these ratios is equal to -------- unless x + y + z = 0
(a + b)
Sum of Antecedents
Ans.) As we know, each Ratio = --------------------
Sum of Consequent
x + y y + z z + x (x + y) + (y + z) + (z + x)
-------- = --------- = -------- = ---------------------------
ax + by ay + bz az + bx (ax + by) + (ay + bz) + (az + bx)
(2x + 2y + 2z)
= -----------------------------
(ax + ay + az) + (by + bz + bx)
2 (x + y + z)
= ---------------------------
a (x + y + z) + b (x + y + z)
2 (x + y + z)
= --------------------
(x + y + z) (a + b)
2
= --------- (proven)
(a + b)
a b c
Example.2) If ------- = -------- = --------,
(b + c) (c + a) (a + b)
Sum of Antecedents
Ans.) As we know that, each ratio = ----------------------
Sum of Consequents
a b c a + b + c
So, -------- = -------- = --------- = ------------------------
(b + c) (c + a) (a + b) (b + c) + (c + a) + (a + b)
a + b + c
= ---------------
(2a + 2b + 2c)
(a + b + c) 1
= --------------- = ------
2 (a + b + c) 2
1
= ------- [where (a + b + c) ≠ 0]
2
Now, when a + b + c = 0, we have
b + c = - a, a + b = - c, and a + c = - b
a a
so, -------- = ------ = - 1
b + c - a
b b
now, -------- = ------- = - 1
a + c - b
c c
and, -------- = ------- = - 1
a + b - c
Hence, in this case, each ratio is – 1 (proven)
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