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Math Help - Geometric mean and right triangles

  1. #1
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    Geometric mean and right triangles

    I want to prove the converse of the theorem that: for a right triangle, the altitude from the right vertex is the geometric mean of the segments of the hypotenuse. I am thinking that if I assume |AD|=\sqrt{|BD||DC|}, with D as the foot of the altitude of triangle ABC from A, then \frac {|AD|}{|BD|}=\frac{|DC|}{|AD|}. This would mean that triangles ABD and ADC are similar with shared right vertices at D, by similarity properties and sharing right angles at D because AD is the altitude of the triangle. What I am stuck on is how I can deduce that A is a right vertex. I was thinking that the altitudes of the smaller right triangles originating from D would be parallel to the lines extending from the angle at A but since they would be perpendicular to those lines, this would have to imply that A is a right vertex.

    Thanks for the help.
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  2. #2
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    Actually, I just realized the converse is false if the foot of the altitude at D is outside the segment BC. I thought of a triangle with an obtuse angle.
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