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In the following configuration, $PQR$ is the orthic triangle of $ABC$: enter image description here

I have to prove (or find) different things:

  1. The sides of the orthic triangle are antiparallel to the sides of $ABC$
  2. The orthocenter of $ABC$ is the incenter of $PQR$
  3. Which points are $A,B,C$ with respect to $PQR$?

I would be glad to receive some hints.

  • 0
    how are these triangles related? – 2017-02-08
  • 0
    You need a figure here. – 2017-02-08
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    PQR(DEF) is the orthic triangle of ABC. Added a figure to the bottom of the post. Note: Changed DEF to PQR to match the image. – 2017-02-08

1 Answers 1

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In order to prove 1., notice that $APQC$ is a cyclic quadrilateral and so on.
To prove 2., perform some angle chasing.
About 3.: if $H$ (the orthocenter of $ABC$) is the incenter of $PQR$, it follows that $A,B,C$ are the excenters of $PQR$. It is interesting to point out that 2. implies that the orthic triangle is the cevian triangle with the shortest perimeter.