Let be a triangle such that , and . Find the area of the triangle.
Since it follows from the law of sines that
Hence . Thus,
To completely justify the title of the post we give another solution based on the following proposition:
Proposition: Let be a given triangle such that . Prove that
Proof: We are working on the following figure.
Let be the bisector of . Then:
and the result follows.
Now, the area of the initial triangle is given by Heron’s formula.