This is clearly false. For example, consider $AB = 3, AC = 4, BC = 5$, and $D$ be the foot of the altitude from $A$ to $\overline{BC}$.

MP8148 wrote: This is clearly false. For example, consider $AB = 3, AC = 4, BC = 5$, and $D$ be the foot of the altitude from $A$ to $\overline{BC}$. Ok but it isn't unique. Tricky. "Unique" being the key word. Let $\omega$ be the circumcircle of $ABC$. Extend $AD$ to intersect $\omega$ at $T$. We know $TD \cdot DA= DB\cdot DC=DA^2$ thus $AD=DT$. Let the parallel from $T$ to $BC$ intersect $\omega$ at $K$. $BC$ bisects $AK$ (in lets say $D'$). Obviously $AD' \cdot D'K = BD' \cdot D'C$ and we get $AD'^2=BD' \cdot CD'$ thus we have two points $D$ and $D'$ which fulfil the given statement. Thus we must have $D=D'$ to have a unique point. So we have $(A,T,K)$ are collinear. Equivalently $ABC$ is isosceles, and we get $DA=DB=DC$ and by Pythagoras we get $AB=AC=\frac{BC\sqrt{2}}{2}$, and we get $AB+AC= \sqrt{2}\cdot BC$

@above It is not necessarily true that ABC is isosceles. Full solution: After you get that T is midpoint of the arc BC, appyling Stewart theorem in BTC you get that DT = TC * sqrt(2) and now Ptolemys on ABTC finishes the problem

Romania JTST 2018