Geometry Problem

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

Click the figure below to see the complete problem 842.

## Wednesday, January 2, 2013

### Problem 842: Triangle, Medians, Centroid, Four Circumcenters, Perpendicular, Congruence, Similarity

Labels:
centroid,
circumcenter,
congruence,
median,
perpendicular,
similarity,
triangle

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By Problem 841,

ReplyDeleteHM=BE/2, HN=AD/2

∴ HO4×HM=HO2×HN

⇒ HO4×BE/2=HO2×AD/2

⇒ HO4×BE=HO2×AD

From problem 841, HM=BE/2 ; HN=AD/2 and HO4*HM = HO2*HN

ReplyDeleteIt simply leads to the result HO4*BE = HO2*AD

Note that the result is true for any point G inside triangle ABC.

ReplyDeletePeter Tran