My Math Forum prove that angle formed at orthocenter is supplement of the angle formed at the vertx

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 November 16th, 2015, 06:13 PM #1 Newbie   Joined: Nov 2015 From: India Posts: 14 Thanks: 1 prove that angle formed at orthocenter is supplement of the angle formed at the vertx I was searching for properties of orthocentre, circumcentre, incentre etc., and I came across this statement: the angle formed at the orthocenter is the supplement of the angle at the vertex (/_BAC + /_BHC = 180°). (Here ABC is a triangle where A is the vertex I am talking about and H is the orthocentre.) Can someone kindly prove it? Thanks for your help. Last edited by skipjack; November 17th, 2015 at 05:16 AM.
November 17th, 2015, 03:26 AM   #2
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Quote:
 Originally Posted by randomgamernerd I was searching for properties of orthocentre, circumcentre, incentre etc., and I came across this statement: the angle formed at the orthocenter is the supplement of the angle at the vertex (/_BAC + /_BHC = 180°). (Here ABC is a triangle where A is the vertex I am talking about and H is the orthocentre.)
Just chase angles around the picture. Here is one way:

$\displaystyle \angle BHC=90+ \angle HCA = 90 + 90 - \angle BAC$

Last edited by skipjack; November 17th, 2015 at 05:17 AM.

 November 17th, 2015, 08:03 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 The total of the interior angles in any quadrilateral add to 360 degrees. And, here, two of the angles are right angles.

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