Name and proof of the identity $c=\frac{a_1}{b_1}=\frac{a_2}{b_2}$ then $c=\frac{a_1+a_2}{b_1+b_2}$

Question

I was shown in a textbook (though not a mathematics one) the following identity:

If

$$c=\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=\dots=\frac{a_n}{b_n}$$ then

$$c=\frac{a_1+a_2+a_3+\dots+a_n}{b_1+b_2+b_3+\dots+b_n}$$

Does this identity have a name and what is the proof for it?

Answer 1

It has a name:
"Theorem on Equal Ratios"
look here

Proof:$$c=\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=…=\frac{a_n}{b_n}$$ Substitute $a_i$=$c\cdot b_i$
i.e. $a_1=c\cdot b_1$
$a_2=c\cdot b_2$
so on...

Consider: $$ \frac{a_1 + a_2 + .... + a_n}{b_1 + b_2 +....+ b_n}$$ Substitute for all $a_i$
$$ \frac{b_1 \cdot c + b_2 \cdot c + .... + b_n \cdot c}{b_1 + b_2 +....+ b_n}$$

Take c common from numerator:
$$ \frac{(b_1 + b_2 +....+ b_n)\cdot c}{b_1 + b_2 +....+ b_n}$$
Did you get c back?

So, $$c= \frac{a_1 + a_2 + .... + a_n}{b_1 + b_2 +....+ b_n}$$

Answer 2

Let $$c=\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=…=\frac{a_n}{b_n}=t$$ then we get $$\frac{b_1t+b_2t+b_3t+...+b_nt}{b_1+b_2+b_3+...+b_n}=t=c$$