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In triangle ABC, angle A = 2 * angle B.  Prove that the side lengths satisfy a^2 = b(b + c).

Jan 19, 2020

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In triangle ABC, angle A = 2 * angle B.

Prove that the side lengths satisfy a^2 = b(b + c).

sin-rule

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{\sin(A)}{a}} &=& \mathbf{\dfrac{\sin(B)}{b}} \quad & | \quad A=2B \\\\ \dfrac{\sin(2B)}{a} &=& \dfrac{\sin(B)}{b} \quad & | \quad \sin(2B)=\sin(B)\cos(B) \\\\ \dfrac{\sin(B)\cos(B)}{a} &=& \dfrac{\sin(B)}{b} \quad & | \quad :\sin(B) \\\\ \dfrac{\cos(B)}{a} &=& \dfrac{1}{b} \\\\ \mathbf{\cos(B)} &=& \mathbf{\dfrac{a}{2b}} \\ \hline \end{array}$$

cos-rule

$$\begin{array}{|rcll|} \hline \mathbf{b^2} &=& \mathbf{a^2+c^2-2ac\cos(B)} \quad & | \quad \mathbf{\cos(B)=\dfrac{a}{2b}} \\ b^2 &=& a^2+c^2-2ac\dfrac{a}{2b} \\ b^2 &=& a^2+c^2-\dfrac{a^2c}{b} \\ \mathbf{\dfrac{a^2c}{b}} &=& \mathbf{a^2+c^2-b^2} \\ \hline \end{array}$$

cos-rule

$$\begin{array}{|rcll|} \hline \mathbf{a^2} &=& \mathbf{b^2+c^2-2bc\cos(2B)} \quad & | \quad \cos(2B)=\cos^2(B)-\sin^2(B) = 2\cos^2(B)-1 \\ a^2 &=& b^2+c^2-2bc\left(2\cos^2(B)-1\right) \quad & | \quad \mathbf{\cos(B)=\dfrac{a}{2b}} \\ a^2 &=& b^2+c^2-2bc\left(2\dfrac{a^2}{4b^2}-1\right) \quad & | \quad \mathbf{\cos(B)=\dfrac{a}{2b}} \\ a^2 &=& b^2+c^2-\dfrac{a^2c}{b}+2bc \quad & | \quad \mathbf{\dfrac{a^2c}{b}=a^2+c^2-b^2} \\ a^2 &=& b^2+c^2-(a^2+c^2-b^2)+2bc \\ a^2 &=& b^2+c^2-a^2-c^2+b^2+2bc \\ a^2 &=& b^2 -a^2 +b^2+2bc \\ 2a^2 &=& 2b^2 +2bc \quad & | \quad : 2 \\ a^2 &=& b^2 +bc \\ \mathbf{a^2} &=& \mathbf{b(b +c )} \\ \hline \end{array}$$

Jan 20, 2020