Problem: The side $BC$ of the triangle $ABC$
is extended beyond $C$ to $D$ so that $CD = BC$.
The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$.
Prove that if $AD=BE$ then the triangle $ABC$ is right-angled.

Solution: Let ray $DA$ meet $\ol{BE}$ at $M$.
Consider the triangle $EBD$.
Since the point lies on median $\ol{EC}$, and $EA = 2AC$,
it follows that $A$ is the centroid of $\triangle EBD$.

\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(180);
pair C = dir(0);
draw(A--B--C--cycle, black+1.4);
pair D = 2*C-B;
pair E = 3*A-2*C;
draw(C--D);
draw(A--E);
draw(B--E, dotted);
draw(A--D, dotted);

pair M = extension(A, D, B, E);
draw(B--E--D--M);

dot("$A$", A, dir(A));
dot("$B$", B, dir(-90));
dot("$C$", C, dir(-90));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$M$", M, dir(M));

/* TSQ Source:

A = dir 50
B = dir 180 R-90
C = dir 0 R-90
A--B--C--cycle black+1.4
D = 2*C-B
E = 3*A-2*C
C--D
A--E
B--E dotted
A--D dotted

M = extension A D B E
B--E--D--M

*/
\end{asy}
\end{center}


So $M$ is the midpoint of $\ol{BE}$.
Moreover $MA = \frac 12 AD = \frac 12 BE$; so $MA = MB = ME$ and hence $\triangle ABE$ is inscribed in a circle with diameter $\ol{BE}$.
Thus $\angle BAE = 90\dg$, so $\angle BAC = 90\dg$.