I don't understand why do you say "I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle", because, to me at least, Pythagoras tells me that it is a right angle.
To see this, you must know that Pythagoras' theorem actually is an "if and only if". That is, if you have a triangle with two sides named $u$ and $v$, then the third one is $u-v$ and Pythagoras says:
$$
u\cdot v = 0 \qquad \Longleftrightarrow \qquad \| u \|^2 + \|v\|^2 = \|u-v\|^2 \ .
$$
That is, if sides (cathetuses) $u$ and $v$ are perpendicular to each other, then the sum of their squares is equal to the square of the third side (hypotenuse) $u-v$. But the reciprocal is also true, if the equality on the right hand side of the displayed formula above holds, then $u$ and $v$ are perpendicular to each other.
The proof is easy if you have already seen some Linear Algebra (if not, you may skip what follows, and go for the actual computations for your problem down below). So, the proof:
$$
\|u\|^2+ \|v\|^2 - \|u-v\|^2 = u\cdot u + v\cdot v - (u-v)\cdot (u-v) = -2u\cdot v \ .
$$
As for your problem, let's compute the three sides:
$$
\|be\| = \sqrt{3^2 + 3^2} = \sqrt{18} \ , \quad \|ce\|= \sqrt{2^2 + 2^2} = \sqrt{8} \quad \text{and} \quad \|bc\|= \sqrt{1^2 + 5^2} = \sqrt{26} \ .
$$
And the quantities appearing in the right hand side of the "extended" Pythagoras' theorem:
$$
\|be\|^2 + \|ce\|^2 = 18 + 8 = 26 \qquad \text{and} \qquad \|bc\|^2 = 26 \ .
$$
Hence, $be$ and $ce$ are perpendicular to each other.