Two lines $p$ and $q$ intersect at $X$. The angle between lines $p$ and $q$ is $45^\circ$. Let $A$ be a point inside the $45^\circ$ angle formed by $p$ and $q$. Let $B$ be the reflection of $A$ through line $p,$ and let $C$ be the reflection of $A$ through line $q.$ Find $\angle BXC,$ in degrees.
Angle BDX = 90 = angle ADX
DA = BD
XD = XD
So, by LL, triangles BXD and AXD are congruent right triangles
And by the same reasoning triangles, AXE and CXE are congruent right triangles
And angle DXE = the angle formed by the intersection of lines p,q = 45 = the sum of angles AXD and AXE
But angles BXD and CXE = angles AXD and AXE....so their sum is also = 45
So angle BXC = AXD + AXE + BXD + CXE = 45 + 45 = 90