Geometry Problem

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

Click the figure below to see the complete problem 670.

## Saturday, September 17, 2011

### #Geometry Problem 670: Triangle, 60 Degrees, Orthocenter, Incenter, Circumcenter, Congruence

Labels:
60 degrees,
circumcenter,
congruence,
incenter,
orthocenter,
triangle

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http://img835.imageshack.us/img835/1895/problem670.png

ReplyDeletePer the result of problem 662

We have ( AHC)=(AIC)=(AOC)=120

And 5 points A, H, I, O and C are co cyclic ( see picture)

Since (B)=60 >> (BAH)=30

Since (AOC)=120 >> (OAC)=30

AI is and angle bisector of (BAC) >> AI is also an angle bisector of (HAO)

Arc HI=arc IO and IH=IO

Peter Tran

B = 60°, A + C = 120°.

ReplyDeleteLet D be the midpoint of BC

BH = 2 OD = 2R cos B = R = BO

∠HBA = 90° - A

∠OBC = ∠OCB = 90° - ∠BOC/2

= 90° - A = ∠HBA

Already BI bisects ∠ABC

Follows BI bisects ∠HBO and

∆BIH ≡ ∆BIO

∴HI = IO

Let D be midpoint of BC and let CH extended meet AB at E.

ReplyDeleteBE = a/2 = BD and < ABH = < OBD = 90-A

Hence tr.s BHE and BDO are congruent ASA and so BH = BO.

Now < HBI = < OBI = B/2 - (90-A)

Hence tr.s HBI and OBI are congruent SAS

Therefore HI = OI

Sumith Peiris

Moratuwa

Sri Lanka