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## Exploring Analytical Geometry with Mathematica / Edition 1

View Metrics. Email alerts New issue alert. Cosets Pages Hibbard, Allen C. Preview Look Like? Groupoids Pages Hibbard, Allen C. Ringoids Pages Hibbard, Allen C. Morphoids Pages Hibbard, Allen C. Show next xx.

Read this book on SpringerLink cover old. Recommended for you. Hibbard Kenneth M.

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PAGE 1. This is Corollary 3 of Ceva's theorem. Thus, the fact that, in a triangle, angle bisectors are concurrent, implies the fact that altitudes in a triangle are also concurrent.

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The other two angles are treated similarly. In other words, the medians are perpendicular to the sides and, therefore, coincide with the altitudes.

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The altitudes then intersect at the centroid of the triangle which is obviously equilateral in this case. A proof in the circular coordinates leads directly to the Euler line and a nice theorem by J. Two short proofs of which the second is the clearest proof I ever came across.

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For example,. This is very much like the first proof by Vector Algebra, but we start with an identity that is valuable in its on right.

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The identity is attributed to L. Thanks to Bianco for this proof. See also Altshiller-Court's College Geometry , p. Vladimir Zajic. Two lines none parralel to any coordinate axis are perpendicular to each other if and only if the product of their tangents is equal to -1 minus one.

Again, we have to set up only 1 equation, the other 2 are given by the cyclic permutation of A, B, C. Since the solution is invariant with respect to the cyclic permutation of A, B, C, it follows that the same coordinates x O and y O are solution of any two altitude coordinates and the 3 altitudes indeed intersect in a single point.