Geometry Problem 1386: Thabit ibn Qurra (826 -901) Theorem and more conclusions, generalized Pythagorean Theorem to any triangle

Proposition

The figure below shows a triangle ABC of sides a, b, c with the cevians BM and BN so that the angles A and C are congruent to the angles NBC and ABM, respectively. BM = d, BN = d1, AM = m, and CN = n. Prove that  (1) triangle MBN is isosceles: d = d1;  (2) triangles AMB , BNC and ABC are similar;  (3) a2 = b.n, similarly c2 = b.m;  (4) d2 = m.n;  (5) a2 + c2 = b.(m+n) Qurra's theorem;  (6) 1/a2 + 1/c2 = (m+n)/(b.m.n).
 

Geometry Problem 1386: Thabit ibn Qurra Theorem and more conclusions, generalized Pythagorean Theorem to any triangle


Key Term / Hints Description
Thabit ibn Qurra's Theorem A generalization of the Pythagorean Theorem to any triangle, involving cevians and angle congruences.
Triangle ABC The primary triangle in the problem with sides a, b, and c.
Cevian A line segment from a vertex to the opposite side of the triangle, intersecting at a point.
Isosceles Triangle A triangle with at least two equal sides.
Similarity of Triangles Triangles that have the same shape but not necessarily the same size, meaning their corresponding angles are equal, and their corresponding sides are proportional.
Generalized Pythagorean Theorem The relationship among the sides of a triangle and cevians, extended from the classical Pythagorean theorem.
Proposition Results Results including the conditions for isosceles triangles, similar triangles, and specific algebraic relationships between sides and cevians.

Geometric Art: Hyperbolic Kaleidoscope of problem 1386

Geometric Art: Hyperbolic Kaleidoscope of problem 1386

See also
Conformal Mapping or Transformation of Problem 1386


Flyer of problem 1386 using iPad Apps

Flyer of Geometry Problem 1386 Thabit ibn Qurra Theorem and more conclusions, generalized Pythagorean Theorem to any triangle using iPad Apps, Tutor