Let ABC be a triangle. We construct squares ABST and ACUV with centers Osub1 and Osub2, respectively, as shown. Let M be the midpoint

Here's  (a)

Draw BV  and TC

In triangles TAC and BAV,  TA  = BA, AC  = AV

Angle TAB  = angle VAC  and angle BAC  = angle BAC

So....angle TAB  + angle BAC  = angle VAC  + angle BAC

So...angle TAC  = angle VAB

So by SAS, triangle TAC  is congruent to triangle BAV

So  TC  = BV

Now.....call the intersection of  BV and TC, N

And call the intersection of  AC and BV, O

Note that  angle TCA  = angle BVA

And angle AOV  = angle NOC

So.....by AA similarity, triangle AOV  is similar to triangle NOC

But angle OAV  is right.....so...angle ONC  is also right.....

Thus.....TC  is perpendicular to BV