Famous theorems in plane geometry and geometry in space

Pythagorean Theorem, Ceva Theorem, the 3 orthogonal relationship Van Aubel

Pythagorean Theorem

In a rectangular triangle, the square of the hypotenuse is equal to the length of the squares of the cathetes.

ΔABC rectangular triangle having the measure of the A angle equals 90°

Then we have:
BC2=AB2 + AC2

 

 

 

 

Reciprocal of Pythagorean Theorem

If in a triangle the square of the length of one side is equal to the sum of the squares of the lengths of the other two sides then the triangle is rectangular.

If in ΔABC we have BC2=AB2+AC2 then m(A)=90°.

 

The theorem of three perpendiculars

Let  α a geometric plan, (d) a right included in the geometric plan α and A ∉ α, O ∈ α, O ∉ d. If AO⊥α, OB ⊥d, then AB⊥d.

The 3 orthogonal Theorem

 

The reciproc 1. Let α a geometric plan, (d) a right line included in the plan α, A ∉ α, O ∈ α, O ∉ d, B∈d. If AOα and ABd, then OBd.

The reciproc 2. Let α a geometric plan, (d) a right line included in the plan α, A ∉ α, O ∈ α, O ∉ d, B∈d. If AOOB, ABd and OBd, then AOd.

 

Ceva's Theorem

Let be the triangle ABC and the points A'∈(BC), B'∈(CA), C'∈(AB). If the right lines AA', BB' and CC' are concurrent (htey have a common point), then we have:

Ceva's Theorem relation formulas

 

Ceva Geometric Theorem

The reciprocally of Ceva's Theorem. Let be the triangle ABC and the points A'∈(BC), B'∈(CA), C'∈(AB). If the bellow relation is true

Ceva Theorem geometry

then the right laines AA', BB' and CC' are concurrent (have a common point).

 

Van Aubel relationship

Let the triangle ABC and the points A'(BC), C∈(AC'), B∈(AB'). If the right lines AA', CB' È™i BC' are concurrent into a point P, then exists the relation:

van aubel relation geometry

 van aube geometric figure with a triangle

 

 

Menelaos Theorem

Let the triangle ABC and A', B' and C' three points such as  A'∈(BC), B'∈(CA), C'∈(AB). If the points A', B' and C' are collinear, then:

Menelaos theorem geometry

Menelaos theorem Geometry triangle

Menelaos reciprocically. Let the triangle ABC and the points A'∈(BC), B'∈(CA), C'∈(AB) such as two of the points A', B', C' are situated on two sides of the triangle, and the 3th on the extension of the 3th side of the triangle. If it is true that

Menelaos reciproc

then the points A', B' and C' are collinear.

 

Theorem parallels unequidistant


Several parallel lines on two secant  determined some proportionate segments. Let d1 || d2 || d3 coplanare with the right lines a and b and

, ,

whatever k ∈ {1, 2, 3}. Then:

 

 

Theorem bisecting

In a triangle the bisector of the angle divides the outer face of a side opposite to the side that is proportional to the segments that form the angle.

bisecting theorem geometryIn the triangle ABC let the bisect (AD of an angle BAC, with D[BC]. Then we have

theorem relation

 

 

 

 

 

Thales' Theorem

Let the triangle ABC and the line d || BC, D and E intersections point of the line d with the sides AB and AC. Then:

Thales theorem geometry   or   Thales formula

 

Thales teorem geoemtry triangle figure

 

 

Keywords: Theorem, Pythagoras, geometry, Menelaus, Ceva's Theorem, Theorem of the 3 perpendicular relationship Van Aubel

 

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