Prove that, in a triangle, $α = 2β \ \implies \ a^2 = b^2 + bc $

Question

I would be very grateful if someone could help me to solve this task!

I have to prove that in a triangle ABC, if $α = 2β$, then $a^2 = b^2 + bc $

Thank you in advance!

Answer 1

Sine rule:

$$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta} \rightarrow \frac{a}{2\sin \beta \cos \beta}=\frac{b}{\sin \beta} \rightarrow a=2b\cos \beta$$

Cossine rule:

$$b^2=a^2+c^2-2\cdot a\cdot c\cdot\cos \beta \Rightarrow b^2=a^2+c^2-2\cdot a\cdot c \cdot\frac{a}{2b}$$

$$b^3=a^2b+bc^2-a^2c \Rightarrow b(b^2-c^2)=a^2(b-c)$$

$$(b-c)[a^2-b(b+c)]=0$$

So,

$b=c$ or $a^2=b^2+bc$

If $b=c$ then $a=b\sqrt{2}$ then we get a right triangle and still have the relation $a^2=b^2+bc$.

Answer 2

Since $A=2B$

$$ \sin(A-B)=\sin(B) $$ $$ \sin(A+B)\sin(A-B)=\sin B\sin C $$ $$ \sin^2A-\sin^2B=\sin B\sin C $$ Applying the sine rule we get

$ a^2-b^2=bc $