Given an angle $\angle{APB}$ and a point $P' \neq P$ on its external bisector, prove that $AP + BP < AP' + BP'$.
My first approach was to try Pythagoras, constructing points $A'$ and $B'$ orthogonal projections of $A$ and $B$ respectively onto the external bisector, and since $AA'$ and $BB'$ are perpendicular to the bisector, then triangles $AA'P'$ and $BB'P'$ are right triangles. But I got to a dead end because Pythagoras uses squares of sides and I need to prove a linear inequality.
My second idea was to notice that if we define an ellipse with foci $A$ and $B$ and a point $P$, and another ellipse with foci $A$ and $B$ and a point $P'$, then it is sufficient to prove that the semiaxis of the latter ellipse is greater than the former, since the equation of an ellipse is $AP + BP = k$. But I have not been able to continue from there.
Best Answer
This problem can be done with or without an ellipse.
$(1)$ Using ellipse
With $A, B$ as foci and $P$ a point on the ellipse, the external angle bisector is just the tangent at $P$.
Since the tangent lies outside the ellipse, it is easy to prove that $$AP+BP \lt AP'+BP'.$$
$(2)$ Without using ellipse.
Extend $AP$ to $B'$ such that $PB'=PB$ and join $P'B'$, $P'B$ as shown.
$\Delta PP'B \cong \Delta PP'B' \implies BP'=B'P'$.
Thus
\begin{align} AP'+BP' &=AP'+B'P' \\ & \gt AB' \\ &= AP+PB' \\ &= AP+BP \end{align}