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November 16th, 2015, 06:13 PM   #1
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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.
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November 17th, 2015, 03:26 AM   #2
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Quote:
Originally Posted by randomgamernerd View Post
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.
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November 17th, 2015, 08:03 AM   #3
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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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