If the perimeter of a the right-angle triangle is six times its smallest side,
find the ratios of the three sides.
I tried solving it by using the normal area and volume.
[Math] Find the ratios of the sides of a triangle
trianglestrigonometry
Related Solutions
As an argument for the uniqueness of C, we can use sine rule. With your notation, $$\frac{\sin|\angle ACB|}{|AB|}=\frac{\sin|\angle BAC|}{BC}$$ As $AB$ is the shortest side, it is opposite the smallest angle. In particular, we must have that $ACB$ is acute, so sine rule uniquely determines it. So we can then determine all the angles, and all the sides.
Your construction can produce two choices for C if $|\angle BAC|\leq \frac{\pi}{2}$. But one of the points will give $|AB|\geq |AC|$, so won't give the prescribed triangle.
A much easier solution would be to start by recognising that the center of the incircle is the intersection of the internal angle bisectors of the triangle. The triangle is a $120-30-30$ degree triangle, and by rather obvious symmetry, $\angle BEA = 90^{\circ}$.
That gives $AE = \sqrt 3\cot 15^{\circ}$ and therefore the base of the triangle is twice that.
The height $BE = AE \tan 30^{\circ} = \sqrt 3\cot 15^{\circ}\tan 30^{\circ}$
So the area is $\frac 12 (2)\sqrt 3\cot 15^{\circ}\sqrt 3\cot 15^{\circ}\tan 30^{\circ} = 3\cot^2 15^{\circ}\tan 30^{\circ} \approx 24.124$.
Best Answer
Start by making three lengths $a$, $b$, and $\sqrt{a^2+b^2}$. Add these and set equal to $6a$ and solve for $b$ in terms of $a$. I'll leave the rest up to you.