Given the triangle $\rm \triangle ABC$, it is known that $\rm \hat A = 60^\circ < \hat B$. The circumference centered at $\rm B$ with radius $\rm BC$
intersects $\rm AC$ again at the point $\rm D \neq C$. Prove that $\rm AB + AD = AC$.
Proposed by Andrés Sáez Schwedt on behalf of the RSME, as the junior problem of the month (jun. 2021).
Most of the times, sketching the construction helps noticing useful insights that are not clear from the plain description of the picture, as it may be seen
in the right-hand side.
In this case, because of $\rm C$ and $\rm D$ lie in the same circumference, we have that ${\rm BC = BD} = r$. Furthermore, because $\rm \angle DAB$ and $\rm \angle BAC = \hat A$ are supplementary angles, they add up to $180^\circ$, so $\rm \angle DAB = 120^\circ$.
Knowing the previous two facts motivates the use of the law of cosines, because it will probably help relate $\rm AB$, $\rm AC$ and $\rm AD$ in some sort; indeed $$ \begin{align*} \rm BC^2 &= \rm AB^2 + AC^2 - AB\ AC,\\ \rm BD^2 &= \rm AB^2 + AD^2 + AB\ AD. \end{align*} $$ Since the two leftmost expressions are identical – as it has already been discussed – it follows that $$ \begin{align*} &\rm \cancel{AB^2} + AC^2 - AB\ AC = \cancel{AB^2} + AD^2 + AB\ AD\\ \implies &\rm \cancel{(AC + AD)}(AC - AD) = AB\cancel{(AC + AD)}\\ \implies &\rm AC = AB + AD; \end{align*} $$ just as we wanted to prove.
In this case, because of $\rm C$ and $\rm D$ lie in the same circumference, we have that ${\rm BC = BD} = r$. Furthermore, because $\rm \angle DAB$ and $\rm \angle BAC = \hat A$ are supplementary angles, they add up to $180^\circ$, so $\rm \angle DAB = 120^\circ$.
Knowing the previous two facts motivates the use of the law of cosines, because it will probably help relate $\rm AB$, $\rm AC$ and $\rm AD$ in some sort; indeed $$ \begin{align*} \rm BC^2 &= \rm AB^2 + AC^2 - AB\ AC,\\ \rm BD^2 &= \rm AB^2 + AD^2 + AB\ AD. \end{align*} $$ Since the two leftmost expressions are identical – as it has already been discussed – it follows that $$ \begin{align*} &\rm \cancel{AB^2} + AC^2 - AB\ AC = \cancel{AB^2} + AD^2 + AB\ AD\\ \implies &\rm \cancel{(AC + AD)}(AC - AD) = AB\cancel{(AC + AD)}\\ \implies &\rm AC = AB + AD; \end{align*} $$ just as we wanted to prove.