Rectangle $WXYZ$ is drawn on $\triangle ABC$, such that point $W$ lies on segment $AB$, point $X$ lies on segment $AC$, and points $Y$ and $Z$ lies on segment $BC$, as shown. If $m\angle BWZ=26^{\circ}$ and $m\angle CXY=64^{\circ}$, what is $m\angle BAC$, in degrees? [asy]
unitsize(0.3cm);
draw((0,0)--(4.5,6)--(21,0)--cycle);
draw((3,0)--(3,4)--(10,4)--(10,0)--cycle);
label("B",(0,0),SW);
label("W",(3,4),W);
label("A",(4.5,6),N);
label("X",(10,4),NE);
label("C",(21,0),E);
label("Y",(10,0),S);
label("Z",(3,0),S);
label(scale(0.7)*Label("$26^{\circ}$",(2.4,2.6),S));
label(scale(0.8)*Label("$64^{\circ}$",(11,3.5),S));
[/asy]
Since $WXYZ$ is a rectangle, angles $XYC$ and $WZB$ are right angles. Since the acute angles of a right triangle sum to $90^\circ$, $m\angle WBZ=90-m\angle BWZ=90-26=64^\circ$ and $m\angle XCY=90-m\angle CXY=90-64=26^\circ$. In triangle $ABC$, the interior angles must sum to $180^\circ$, so $m\angle BAC=180-m\angle WBZ-m\angle XCY=180-64-26=\boxed{90^\circ}$.