Find $\angle B$ for Triangle $ABC$ with Bisectors $AD,BE$

[maxkor]

Let $ABC$ be a triangle with $\angle A= 60^{\circ},$ and $AD,BE$ are bisectors of $A,B$ respectively where $D\in BC, E\in AC.$ Find the measure of $B$ if $AB+BD=AE+BE.$

 

[jvicens]

Hello, thank you for your question. First, we can use the Angle Bisector Theorem to find the lengths of $AD$ and $BE$. Since $AD$ and $BE$ are bisectors, we know that $\frac{BD}{DC} = \frac{AB}{AC}$ and $\frac{BE}{AE} = \frac{AB}{BC}$. Since $\angle A = 60^{\circ}$, we can use the fact that the angles in a triangle add up to $180^{\circ}$ to find that $\angle B = 90^{\circ}$.

Next, we can use the given information to set up an equation. Since $AB+BD=AE+BE$, we can substitute in the lengths we found using the Angle Bisector Theorem to get $\frac{AB \cdot BC}{AC} + \frac{AB \cdot AC}{BC} = \frac{AB \cdot BC}{AE} + \frac{AB \cdot AE}{BC}$. Simplifying this equation, we get $AB^2 = AE \cdot AC$.

Using the Law of Cosines, we can find the length of $AB$ in terms of $B$. Since $\angle A = 60^{\circ}$, we have $AB^2 = AC^2 + BC^2 - 2AC \cdot BC \cdot \cos(60^{\circ})$. Simplifying this, we get $AB^2 = AC^2 + BC^2 - AC \cdot BC$.

Now, we can substitute this into our previous equation to get $AC^2 + BC^2 - AC \cdot BC = AE \cdot AC$. Factoring out $AC$, we get $AC(AC-BC) = AE \cdot AC$. Since $AC \neq 0$, we can divide both sides by $AC$ to get $AC-BC = AE$.

Finally, we can use the Law of Cosines again to find the measure of $B$. Since $\angle A = 60^{\circ}$, we have $AC^2 = AB^2 + BC^2 - 2AB \cdot BC \cdot \cos(60^{\circ})$. Substituting in the lengths we found earlier, we get $AE^2 = AB^2 + BC^2 - AB \cdot BC$. Simplifying this, we get $