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Altitudes and Bisectors of a TriangleDate: 05/25/99 at 13:01:28 From: Hanna Subject: Height and bisector in triangle We draw the 3 heights of triangle. Then we connect the meeting points of the heights with the sides of the triangle and get a new triangle. Prove: the heights of the first triangle are bisectors in the new triangle. Date: 05/25/99 at 16:38:13 From: Doctor Floor Subject: Re: Height and bisector in triangle Dear Hanna, Thank you for your question! Let us consider the figure you describe: a triangle ABC, with its orthic triangle A'B'C' made out of the feet of the altitudes (heights) of ABC. The three altitudes pass through a common point H, the orthocenter. We are to show that A'H, B'H, and C'H are the angle bisectors of triangle A'B'C'. Since this will only be true if ABC is acute (if ABC is obtuse, two of the lines A'H, B'H, and C'H become 'external' angle bisectors), I assume your question is limited to such triangles. Here's a picture: Date: 06/06/2002 at 16:01:33 From: Zenon Kulpa Subject: Altitudes and Bisectors of a Triangle Hi, In fact, the construction and all your arguments given later are valid for obtuse triangles as well, and even your drawing is good for that case: just take the original triangle to be ABH (instead of ABC) and then the orthocenter will be at C, while the orthic triangle will remain the same... Best regards, -- Zenon Date: 06/06/2002 at 16:01:26 From: Doctor Floor Subject: Re: Altitudes and Bisectors of a Triangle Hi, Zenon, You are right. When I read the question, I was just thinking of internal angle bisectors. But indeed in the case of an obtuse triangle the internal angle bisectors become external ones, but still angle bisectors. Thanks for the feedback. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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