Inequality between the height of a triangle and the product of the associated bisected lengths on the base

Suppose that $ABC$ is a triangle with base $AB$ and height $XC$ of length $h$, where $X$ is the point of intersection of the height and the base located between $A$ and $B$.

Is the following observation $$(\measuredangle A+ \measuredangle B) \gt \measuredangle C \implies h^2>AX \times XB$$ correct? If so, how to carry out its proof?

P.S. I faced this problem while trying to solve an equality between the sign of the algebraic sum of the angles and the sign of the algebraic sum of squares on the side lengths of a triangle.

In the circle with diameter $AB$, the power of $X$ equals $AX\times XB$. On the other hand, in the corresponding triangle formed by $A, B$ and the point $P$ determined by the power of $X$ lying on the circumference, the sum of the new angles associated with $A$ and $B$ equals $\pi/2$.

In the original triangle, if $\measuredangle A+\measuredangle B > \measuredangle C$ then $\measuredangle A+\measuredangle B > \pi/2$, so that the tip of the height will lie higher than $P$, which concludes the proof.