Let ABC be a triangle. We construct squares ABST and ACUV with centers Osub1 and Osub2, respectively, as shown. Let M be the midpoint of line BC.
(a) Prove that line BV and line CT are equal in length and perpendicular.
(b) Prove that line Osub1 M and line Osub2 M are equal in length and perpendicular.
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