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

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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!

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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   Jun 1, 2018
edited by CPhill  Jun 2, 2018
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Thank you!