# Thread: Geometric mean and right triangles

1. ## 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 $\displaystyle |AD|=\sqrt{|BD||DC|}$, with D as the foot of the altitude of triangle ABC from A, then $\displaystyle \frac {|AD|}{|BD|}=\frac{|DC|}{|AD|}$. 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.

2. 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.