In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter7: Locus And Concurrence
Section7.2: Concurrence Of Lines
Problem 8E: a Is it really necessary to construct all three bisectors of the angles of a triangle to locate its...
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How do i show how this proposition are equivalent to the law of cosines?
In obtuse-angled triangles the square on the side subtending the obtuse angle is
greater than the squares on the sides containing the obtuse angle by twice the
rectangle contained by one of the sides about the obtuse angle, namely that on
which the perpendicular falls, and the straight line cut off outside by the
perpendicular towards the obtuse angle.
Transcribed Image Text:In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle.
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