Proof the triangle is rectangular if the sides and half the circumference form a series

Question

How can I effectively proof the following statement:

When in a triangle, the three sides and half the circumference form 4 consecutive therms of an arithmetic series, proof the triangle is right.

I know that, when working in a right triangle, Pythagoras says that the sides should equal 3x,4x,5x and that half the circumference will therefor always be 12x/2, or 6x. I know this is indeed a series but how do I effectively proof that this always works the other way around?

I tried thinking about the fact that when naming the sides a, b and c b should always be a+x, c=a+2x but how do I translate it in to a proof? What technique or what steps can help me in the right direction?

Answer 1

If $a,b,c$ are sides of triangle and we assume that $\displaystyle a,b,c,\frac{a+b+c}{2}$ is an arithmetic sequence, then we can write \begin{align}a+c&=2b\tag{1}\\b+\frac{a+b+c}{2}&=2c\tag{2}\end{align}

Set $(1)$ in $(2)$ we get $5b=4c$. Set $c=1.25b$ in $(1)$ we get $a=0.75b$, hence sides of triangle are $0.75b,b,1.25b$. Easy to show that $a^2+b^2=c^2$, hence the triangle is right angle.