Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Sunday, November 17, 2019

### Dynamic Geometry Problem 1447: Outer Vecten Point

Labels:
center,
congruence,
dynamic geometry,
equal,
GeoGebra,
perpendicular,
square,
triangle,
Vecten point

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https://photos.app.goo.gl/LavifjVpqeaMipcLA

ReplyDeleteLet S is the midpoint of AB

Connect BF, AI, SO and SQ

Triangle ACI congruent to FCB ( case SAS)

ACI is the rotational image of FCB around C so AI= FB and AI ⊥FB

S, O and Q are the midpoints of AB, AF and BI => SO= SQ and SO ⊥SQ

Triangle SPO congruent to SAQ ( case SAS)

SAQ is the rotational image of SPO around S , 90 degrees

So PO=AQ and PO ⊥AQ

Similarly OQ=CP and OQ ⊥ CP and PQ = BO and BO⊥ PQ

In triangle OPQ, 3 altitudes QA, PC and OB will concur at orthocenter V

The concurrency part is trivial by Jacobi, and the rest part can be done by Baudhiyana's Theorem.

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