# If $\alpha$ is obtuse show that $A_1$ is always on the segment $\left[C' B\right]$

In a problem we have a triangle $[A_1BC]$ and we know that it is obtuse in $A_1$. Also $C'$ is the orthogonal projection of $C$ onto $A_1B$.

The original image does not have the circunference and I've shown the result using properties of angles in a circunference.

In short, it's easy to show that if $A_1$ is not in the segment $\left[C' B\right]$ then the angle will always be acute which contradicts the hypothesis.

Can you find another way of showing this result?

Thank you very much!

Suppose that $A_1$ is not on the segment $C'B$. Then there are two cases.
Case 1: $C'$ is on segment $A_1B$. Then $90^{\circ}=\angle AC'B=\angle C'CA_1+\angle C'A_1C\geq\angle C'A_1C>90^{\circ}$, a contradiction.
Case 2: $B$ is on segment $A_1C'$. Then $\angle CA_1C'+\angle CC'A_1>90^{\circ}+90^{\circ}=180^{\circ}$. But this means the angles of triangle $CA_1C'$ sum up to more than $180^{\circ}$, again a contradiction.