Question

If $a^2+b^2+c^2=ab+ac+bc$ find $\frac{c+a}{b}$

Answer 1

Step 1: Converting the given equation in terms of complete squares

Given $a^2+b^2+c^2=ab+ac+bc$

Now multiplying both sides by 2 and simplifying, we get

$$\begin{gathered} \Rightarrow 2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right) \\ \Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0 \\ \Rightarrow {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2=0 \end{gathered}$$

Step 2: Finding the required value
Since the sum of three square terms is 0, then each square term must be equal to 0

$$\begin{gathered} \therefore a-b=0,b-c=0,c-a=0 \\ \therefore a=b=c \end{gathered}$$

$$\begin{gathered} \therefore \frac{c+a}{b}=\frac{a+a}{a} \\ =2 \end{gathered}$$

Hence the value of $\frac{c+a}{b}$ is $2$