Geometry Problem

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

Click the figure below to see the complete problem 841.

## Wednesday, January 2, 2013

### Problem 841: Triangle, Medians, Centroid, Circumcenters, Perpendicular, Congruence, Similarity

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∵ HO4 perpendicular bisects BG

ReplyDelete∵ MO6 perpendicular bisects GE

∴ HM//BE and HM=BE/2

Similarly, HN=AD/2.

∵ ∠NHO2=∠MHO4=90°

∴ ∠NHO4=∠MHO2

∴ ΔHNO4~ΔHMO2

∴ HM/HO2=HN/HO4

∴ HO4×HM=HO2×HN

Note o2o6 perpendicularly bisect GE ; o3o4 perpendicularly bisect of GB.

ReplyDeleteSay they bisect at points P and Q respectively. So, PQ = BE/2.

Easy observe that HMPQ forms a rectangle. HM = PQ

For the next part, HO4.HM represent double of triangle area of HMO4. And HO2.HN represent double of triangle area of HNO2.

By affine transformation which preserves area, transform the triangle into an equilateral ones, and it is proved