Prove that this is an isoceles triangle

Question

I'm trying to solve a problem here.

It says: "Prove that a triangle is isoceles if $\large b=2a\sin\left(\frac{\beta}{2}\right)$." $B-\beta$ I've tried to prove it but I can't

Can anyone help me?

Answer 1

Square both sides and use the law of cosine: $$b^2 = 4a^2\cdot (sin(B/2))^2 = 2a^2(1 - cosB) = 2a^2(1 - (a^2 + c^2 - b^2)/2ac)$$ . Simplify this equation we get:

$$(a - c)(a^2 - ac - b^2) = 0$$.

Case 1: If $a = c$, we're done.

Case 2: If $a \lt c \Rightarrow a - c \lt 0$, and $a^2 - ac - b^2 = a(a - c) - b^2 \lt 0$, so $LHS \gt 0 = RHS$. Contradiction.

If $a \gt c$, then $a - c \gt 0$ and since$ b^2 < (a - c)^2 ==> a^2 - ac - b^2 > a^2 - ac - (a - c)^2 = a^2 - ac - (a^2 - 2ac + c^2) = ac - c^2 = c(a - c) > 0$, so $LHS > 0 = RHS$. Contradiction again.

So Case 2 can't happen, and so $a = $c and triangle $\hat{ABC}$ is isosceles.

Answer 2

Hint : for the law of sines you have $$ \frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{2a\sin\left(\frac{\beta}{2}\right)}{\sin\beta} \Rightarrow ...$$