Question about angles in a triangle having a median line 
If PS = SR and x = y, is it true that angle QSR must be 90 degrees?
More generally, if PS = SR, is it true that x+y must equal 90?
I have tried drawing circumcircles, expanding the triangle into a parallelogram, and drawing an altitude from S onto QR.  I can't seem to prove QSR is 90.  Which makes me suspect it doesn't have to be.  On the other hand, I can't seem to visualize a triangle (or draw one) where it is clear that PS=SR and x=y, but angle QSR is not 90.  I also feel like using the laws of sines/cosines doesn't prove helpful.
If angle QSR is 90 and PS=SR, x need not equal y.  If angle PQR is 90 and PS=PR, x need not equal y.  If angle PQR is 90 and PS=SR and x=y, then, yes, angle QSR is 90.  There's a lot of combinations of givens and unknowns we can consider.  But for the particular 2 givens, PS=SR and x=y, and the unknown, angle QSR, I seem to be at a loss.  Any thoughts?
EDIT: I have solved this.  The answer to both questions is "no."  For a fixed value of y and a fixed length of QR, x can be written as a function of the length of PR, call it b.  Then taking the example of the fixed values y=30 and QR=2/sqrt(3), as b increases from 2, x strictly decreases from 60 (to 0), and angle QSR decreases from 90 degrees.  Then at some value of b>2 (approximately b=3.1), x will be equal to 30, while QSR will be less than 90.  Thus, there exists a triangle where PS=SR and x=y, but angle QSR is less than 90 degrees.  The same example shows that x and y can both be equal to 30, and thus x+y need not equal 90.
 A: No to everything that you have asked because there are too many free variables to force that in both cases. However, if you add the hypothesis PQR is a right angle then yes to everything you have asked.
It kind of looks like a right angle in the picture but you do not specify it.
You can use the sum of angle formula and law of sines formula to obtain the possible restrictions.
So:
$$QSR + QSP = 180$$
$$RQS + QSR + y = 180$$
$$QPS + QSP + x = 180$$
$$x + y + RQS + QPS =180$$
Gauss Elimination gives:
$$x + QSP + QPS = 180$$
$$y + QSR + RQS = 180$$
$$QSR + QSP = 180$$
The law of sines gives us:
$$sin(x)sin(y) = sin(QPS)sin(RQS)$$
This will leave two free variables. With the extra equation x=y we will have one free variable. And get the extra equation after elimination:
$$RQS - 2QSP - QPS = -180$$
To simplify we can use the product of sines trig identity and then linear substitution to transform the last formula:
$$2sin(x)sin(y) = cos(QPS-RQS) - cos(QPS+RQS) = cos( 180 -x -y - 2RQS ) - cos(180 - x - y)$$
Then in the first case we can go further:
$$2sin^2(x) = cos(180 - 2x -2RQS) - cos(180 -2x) = cos(2x) - cos(2*(x+RQS)) = $$
$$ = 1 - 2sin^2(x) -1 + 2sin^2(x+RQS)$$
Simplifying: $$2sin^2(x) = sin^2(x+RQS)$$
Which has solutions when
$$RQS = π - x - arcsin(sqrt(2) sin(x))$$ in radians.
Letting x be 30 degrees, We get that RQS is 105 degrees and y is forced to be 30 degrees making QSR 45 degrees. But there are infinitely many choices for x that work, it is a free variable in other words.
For the second question 30 degrees plus 30 degrees is 60 degrees not 90 so it is false.
If you enforce the extra condition that you did not mention that PQR is a right angle then the first statement is true which should be clear from the simplified form since
$$sin^2(x+RQS) = sin^2(RPQ) = 1.$$
And for the second statement continuing where we left off.
$$2sin(x)sin(y) = cos(QPS-RQS) - cos(QPS+RQS) = $$
$$= cos( 180 -x -y - 2RQS ) - cos(180 - x - y) = cos( x - y ) - cos(180 - x - y)$$
Simplifying: $$2sin(x)sin(y) = cos (x - y) + cos( x + y ) = 2cos(x)cos(y)$$
Which will happen when $$cos(x + y) = 0$$
Which means that either x+y = 90 degrees or x+y =270 degrees, but the second is too large for the interior of a triangle, and so under the additional constraint the more general question is true.
A: The statement $PS=SR$ means that $S$ is at the midpoint of $PR$, and so $QS$ is a median.
Easy answer first: considering your second "more general" question, observe that $y$ can be obtuse, so the answer is "no", $x+y$ is clearly not always $90^\circ$.
Now the hard one: Taking $x=y$ we immediately have the similarity $\triangle PQR \sim \triangle PQS$ (with the shared $\angle SPQ$). We thus know that $\frac{|PR|}{|PQ|}=\frac{|PQ|}{|PS|}$ and so $|PQ|^2 = |PR||PS| = 2|PS|^2$, so $|PQ| = \sqrt{2\,}|PS|$.
However this doesn't appear to get anywhere in terms of demonstrating a right angle at $Q$ so we will examine for other possibilities.
Consider the line $PR$ with $S$ at its midpoint. Now select a point between $P$ and $S$, say $T$. Now imagine point $Q$ sliding on the line produced from $T$ perpendicular to $PR$. With $Q$ close to $PR$, angle $x$ will be large and angle $y$ will be small. When $Q$ is far from $PR$, angle $x$ will be small and angle $y$ will be large. So somewhere between these limits is a point where $x=y$ for any $T$ between $P$ and $S$. For $T$ at $P$, the same process produces an equal $x$ and $y$. Combined with the relationship above between $|PQ|$ and $|PS|$, we know where this is, so there is a continuum of different triangles that can have this property; not just a right triangle.

Tidying this up in retrospect, it's actually clear that with $S$ at the midpoint of $PR$,
$$ \text{angle }x = \text{angle }y \;\iff \; \triangle PQS \sim \triangle PRQ \;\iff\; |PQ| = \sqrt{2\,}|PS| $$
so the range of possible triangles fulfilling these conditions is illustrated in my gif above.
A: To answer the first question, lets assume that angle QSR is 90 degrees given that x = y and PS = SR. Angle SQR would then be 90 - y, and angle PQR would be 90 - y + x = 90. This means that PQR is a right triangle with the median from Q being an altitude. This would only be possible if the triangle was also isosceles, so QSR can only be 90 degrees if PQR is an isosceles right triangle with right angle Q.
It is clear that all isosceles right triangles will work, but not all triangles with the original conditions are isosceles right triangles. For example, whenever PR = sqrt2 * PQ, the conditions are satisfied but QSR is not 90. Therefore, it is not always true that angle QSR = 90.
For the second question, consider a right triangle this time with right angle at P. It is clear that angles y and PQR sum to 90. However, since angle x comes from the median from Q, it would always be true that angle x is less that PQR, or x + y < 90. Therefore, it is not always true that x + y = 90.
