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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,864

The altitudes of a triangle meet at a point, called the orthocentre.

Proof.

In triangle ABC, mark the midpoints D, E, F of the sides.

Consider triangles AFE and ABC. They have two lines and the included angle in common, so they are similar, with a length scale factor of x2.

So EF is parallel to BC. Similarly for FD//AC and ED//AB.

I have already shown that the perpendicular bisectors of the sides of ABC are concurrent (meet at a point) . See http://www.mathisfunforum.com/viewtopic.php?id=22507

But these lines are the altitudes of DEF. So the altitudes of DEF are concurrent, and O is the orthocentre of DEF.

Will this be true for any triangle?

Label the sides of the triangle D, E and F. Draw lines parallel to each side going through the vertices to produce triangle ABC.

Because of the parallel lines and the common sides AFE, DEF, EDC and FBD are all congruent, so D, E and F are the midpoints of ABC. The result follows.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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