If $\Delta{}ABC$ is a triangle, call the segment perpendicular to $AB$ and containing $C$ the altitude segment at $C$. In brief my question is the following: is it true that for all $0<\theta<\frac{\pi}{2}$ and all triangles $\Delta{}ABC$ satisfying $\measuredangle{}ACB=\theta$, there exists a point $P$ on the altitude segment at $C$ such that $PA+PB+PC<AC+BC$? See the figure below for an example.
Now for some context: I have read in The Secrets of Triangles: A Mathematical Journey by Alfred S. Posamentier and Ingmar Lehmann that if $AB$ is the shortest side of triangle $\Delta{}ABC$ and $P$ is any point in the interior of $\Delta{}ABC$, then $PA+PB+PC<AC+BC$. (The "proof" they give seems rather incomplete.)
While playing with the geometry of the points $A,B,C,P$, I noticed that for many triangles that do not satisfy the condition that $AB$ be the shortest edge of the triangle $\Delta{}ABC$, there nonetheless often exists a point $P$ in the interior of $\Delta{}ABC$ (and especially on the altitude segment at $C$) such that $PA+PB+PC<AC+BC$.
For certain reasons, I am interested in (and am restricting to) the case when $0<\measuredangle{}ACB<\frac{\pi}{2}$ and $P$ is required to lie on the altitude segment at $C$. Is one always guaranteed the existence of such a point $P$ satisfying $PA+PB+PC<AC+BC$ whenever $0<\measuredangle{}ACB<\frac{\pi}{2}$?
Addition: Is it any easier under the further assumption that $\Delta{}ABC$ be acute? This assumption guarantees that the orthocenter lie in the interior of $\Delta{}ABC$. I am quite interested in this sub-case.
If $AB\leq AC$ and $AB\leq BC$ so it's true for any point $P$ inside the triangle.
I hope at least this fact would be completed.
Indeed, let $D\in AC$, $E\in BC$, $P\in DE$ such that $DE||AB$ and $CE=kBC$.
Thus, $CD=kAC$ and by the triangle inequality we obtain: $$PA+PB+PC<AD+DP+BE+EP+PC=AD+BE+DE+PC=$$ $$=(1-k)AC+(1-k)BC+kAB+PC<$$ $$<(1-k)AC+(1-k)BC+kAB+\max\{DC,EC\}=$$ $$=(1-k)AC+(1-k)BC+kAB+k\max\{AC,BC\}=$$ $$=AC+BC+k\left(AB-AC-BC+\max\{AC,BC\}\right)\leq AC+BC.$$