Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.

Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demonstration of problem 892.

## Wednesday, June 26, 2013

### Problem 892: Triangle, 60 degrees, Orthocenter, Circumcenter

Labels:
60 degrees,
angle,
circumcenter,
orthocenter,
triangle

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http://img15.imageshack.us/img15/6545/fyi.png

ReplyDeletelet H’ and O’ are point of symmetry of H and O over AC.

H’ and O’ will be on circle O ….( property of orthocenter and angle AOC= 120)

Due to symmetric property, O and H will be on circle O’, radius=OO’ ( see sketch)

We have Angle(AOO’)=angle(AO’O)=60

And angle(AHO)= ½*angle(AO’O)=30

Extend EC, EC meet the circle at anoter point, T

ReplyDeleteHCF=EBF

HCF=TBE

TBE=EBH && BE Orthogonal to HT => TE=EH

=> AH=AT

ATH=60 => ATH Equilateral triangle

let M midPoint of AT

build HM, HM median and Altitude to AT=> O on HT =>

OH an angle bisector => OHA=30

Problem 892

ReplyDeleteIs <AOC=120=<AHC (<AHC=180-<ABC=180-60=120). Therefore AOHC is concyclic. So

<AHO=<ACB=(180-120)/2=30.