Incenter of a Triangle.

Question

In triangle ABC, bisectors $AA_1$, $BB_1$ and $CC_1$ of the interior angles are drawn. If $\angle ABC=120^ \circ$, what is the measure of $\angle A_1B_1C_1$ ?

I solved this problem as :

Mark D, as the point of intersection for $AA_1$,$BB_1$ and $CC_1$.

The point D is the incenter, that is D is the center of the circle with radii $DA_1$, $DB_1$ and $DC_1$.

So, AB, BC and CA are the tangents to the circle with center D.

So, $\angle DA_1B = DC_1B = 90^\circ$.

So, in quadrilateral $BA_1DC_1$, $\angle A_1DC_1 = 360-90-90-120 = 60^\circ$.

So, in circle center D, $\angle A_1B_1C_1 = \frac{60}{2} = 30^\circ$.

Is this solution correct? ... Please advise.

Answer 1

In triangle $ABC$, using the law of sines and $AA_1$ angle bisector, we get

$$\frac{\text{CA}_1}{A_1B}=\frac{\text{CA}}{\text{AB}}=\frac{\sin 120{}^{\circ}}{\sin C}$$

In triangle $CB_1B$, using the law of sines

$$\frac{\text{CB}_1}{B_1B}=\frac{\sin 60{}^{\circ}}{\sin C}$$

Therefore

$$\frac{\text{CA}_1}{A_1B}=\frac{\text{CB}_1}{B_1B}$$

and therefore $B_1A_1$ bisects angle $BB_1C$

Similarly, $B_1C_1$ bisects angle $AB_1B$. It follows that $B_1A_1$ and $B_1C_1$ are perpendicular and $\measuredangle A_1B_1C_1=90^{\circ}.$

Answer 2

No. It's not correct.

I think the following reasoning is right.

Let $K\in AB$ such that $B$ placed between $A$ and $K$.

Thus, $\measuredangle KBC=60^{\circ},$ which says that $BC$ is a bisector of $\angle KBB_1$ and since $AA_1$ is a bisector of $\angle BAC$, we see that $B_1A_1$ is a bisector of $\angle BB_1C$.

Similarly, $B_1C_1$ is a bisector of $\angle AB_1B$ and $\measuredangle A_1B_1C_1=90^{\circ}.$