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.

Thanks!

AnonymousConfusedGuy
Jun 1, 2018

#1**+2 **

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

CPhill
Jun 1, 2018