I need to prove the existence of an orthocenter for an obtuse triangle.
I tried proving the existence of an orthocenter, meaning a point where the heights of $\triangle ABC$, where $[AB]=c, [AC]=b, [BC]=a$, meet, as following.
- (1) drawing the parallel lines to a $c$ through $C$ and $b$ through $B$ and $a$ through $A$ and so we get a new $\triangle DEF$
- (2) then we get parallelograms and can conclude that $\triangle ABC$ is the inner triangle of $\triangle DEF$
- (3) The points $A, B, C$ cut the sides of $\triangle DEF$ in half.
- (4) The perpendicular bisector of $\triangle DEF$ meets in one point $S$.
- (5) Therefore the heights of $\triangle ABC$ also meet in one point.
Is this a proof? Or doesn't it hold for obtuse triangles?
$\endgroup$ 11 Answer
$\begingroup$The proof is quite true if we assume the concurrency of the perpendicular bisectors (circumcenter of △ $DEF$ ) of the triangle constructed in the argument. The proof no doubt works for all triangles. First we have to state that the altitudes are already drawn and we have to argue about their coincidence with their perpendicular bisectors and hence the concurrency of the circumcenter therefore proving the existence of the orthocenter.
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