A cute problem in Trigonometry or pure plane geometry
Problem.
In a triangle ABC, $latex angle A$ is twice $latex angle B$. Show that $latex a^{2}=b(b+c)$. (In fact, the converse is also true. Prove it!).
Proof.
Method I.
You can use plane geometry also. This is left to you as an exercise.
Method II.
You can use trigonometry also. We may use the sine rule for a triangle to dispose of both the implications simultaneously.
$latex A=2B Longleftrightarrow A-B=B Longleftrightarrow sin{(A-B)}=sin{B} Longleftrightarrow sin{(A-B)}sin{(A+B)}=sin{B}sin{C} Longleftrightarrow sin^{2}{A}-sin^{2}{B}=sin{B}sin{C} Longleftrightarrow (2Rsin{A})^{2}-(2Rsin{B})^{2}=(2Rsin{B})(2Rsin{C}) Longleftrightarrow a^{2}-b^{2}=bc Longleftrightarrow a^{2}=b(b+c)$
