Let us say that we have a circle and a cyclic quadrilateral $ABCD$ where the segment $BD$ passes through the center of the circle, thus being the diameter. Also, $\angle A$ and $\angle C$ are both right angles.
We also have a triangle $DEB$ where the measure of $\angle E$ is $90^\circ$.
The measures of $\angle EBD$ and $\angle DBC$ are $30^\circ$ and $20^\circ$ respectively, making $\angle B$ add up to $50$.
My job was to simply prove that $ABCD$ is a rectangle.
My intuition is that $ABCD$ is definitely a rectangle, since two opposite angles of the quadrilateral are both right angles in the first place. (However, my picky math teacher wants me to even justify this statement.)
This way, I took a different approach, initially proving that $ABCD$ is a parallelogram, thus it is a rectangle. (Since we already have a pair of opposite right angle triangles, where two angles that are left have to be both equal to each other and adding up to $180^\circ$, making $90^\circ$ the only answer.)
However, my end game is to prove that the line $AC$ passes straightly through the center of the circle just like $BD$. In this way, I can't prove that the cyclic quadrilateral is a parallelogram since there is a possibility that $AC$ doesn't pass through the center, contradicting the property that two diagonal lines should bisect each other which only works when $AC$ is a diameter, getting bisected into two radii.
Would this be the case where there is not enough information to perform the proof? I have tried to come up with many different premises and theorems, but in the end, I find myself back to square one.
Currently , $ABCD$ may or may not be Parallelogram , unless we have Extra Details.
Consider these Counter Examples :
Point $A$ could be $A1$ or $A2$ or $A3$ & all Cases will give angle $90^\circ$.
These Quadrilaterals will all satisfy the given Criteria , yet might not be Parallelograms. There may be a Parallelogram among those Points though that is not yet known here.
Your "Picky" teacher was right to ask for Proof for your Claim : The Claim was actually wrong.
ADDENDUM :
I used well-known theorems to conclude that the various angles are all $90^\circ$ : Details are available here :
https://en.wikipedia.org/wiki/Inscribed_angle