Bisectors and equilateral triangle I have the following problem: bisectors $AA',BB'$ and $CC'$ of the triangle $ABC$ intersect the circumcircle in the points $A",B",C".$ Holds the following equivalence: $\frac{AA'}{AA"}+\frac{BB'}{BB"}+\frac{CC'}{CC"}=\frac{9}{4}$equivalent triangle $ABC $ equilateral?
 A: 
In case you requested to prove $$\frac{AA_1}{AA_2}+\frac{BB_1}{BB_2}+\frac{CC_1}{CC_2}=\frac{9}{4} \iff \Delta ABC \text{ is equilateral }$$

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*$ABC \text{ is equilateral }$
This case is quite trivial, as $\frac{AA_1}{AA_2}=\frac{BB_1}{BB_2}=\frac{CC_1}{CC_2}=\frac{r+\frac{r}{2}}{2r}=\frac34$, thus $\frac{AA_1}{AA_2}+\frac{BB_1}{BB_2}+\frac{CC_1}{CC_2}=3\frac34=\frac94$

*$\frac{AA_1}{AA_2}+\frac{BB_1}{BB_2}+\frac{CC_1}{CC_2}=\frac{9}{4}$
Consider second picture:

We can state:
$$BA_2=2R\sin\angle BAA_2=2R\sin\left(\frac{\angle BAC}{2}\right)$$
$$\angle BA_1A_2 = \pi-(A_1BA_2 + BA_2A)=\pi-\frac{\angle BAC}{2}-\angle ACB=\frac{\angle BAC}{2}+\angle ABC$$
Thus by the law of sines you get $$\frac{BA_2}{\angle BA_1A_2}=\frac{A_2A-A_1A}{\sin\left(\frac{\angle BAC}{2}\right)}\\\frac{2R\sin\left(\frac{\angle BAC}{2}\right)}{\sin\left(\frac{\angle BAC}{2}+\angle ABC\right)}=\frac{A_2A-A_1A}{\sin\left(\frac{\angle BAC}{2}\right)}$$
Thus
$$A_2A-A_1A=\frac{2R\sin^2\left(\frac{\angle BAC}{2}\right)}{\sin\left(\frac{\angle BAC}{2}+\angle ABC\right)}$$
and since the length of the angle bisector is known to be $$A_1=\frac{2AB\cdot AC\cos\left(\frac{\angle BAC}{2}\right)}{AB+AC}$$
you can find the ratio $\frac{AA_1}{AA_2}$. Reiterate for the other sides and you get the final expression, where most of the length of the sides will get simplified.
