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With parallel lines, they are congruent. Alternate angles are the four pairs of angles that: have distinct vertex points, lie on opposite sides of the transversal and; both angles are interior or both angles are exterior. If the two angles of one pair are congruent (equal in measure), then the angles of each of the other pairs are also congruent.
The three splitters concur at the Nagel point of the triangle. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter, and each triangle has one, two, or three of these lines. [2] Thus if there are three of them, they concur at the incenter.
In particular it is important to assure that for two given line segments, a new line segment can be constructed, such that its length equals the product of lengths of the other two. Similarly one needs to be able to construct, for a line segment of length , a new line segment of length . The intercept theorem can be used to show that for both ...
Note hatch marks are used here to show angle and side equalities. In elementary geometry the word congruent is often used as follows. [2] The word equal is often used in place of congruent for these objects. Two line segments are congruent if they have the same length. Two angles are congruent if they have the same measure.
Consider a triangle ABC.Let the angle bisector of angle ∠ A intersect side BC at a point D between B and C.The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of the length of side AB to the length of side AC:
Formally, let ABC be a triangle, with arbitrary points A´, B´ and C´ on sides BC, AC, and AB respectively (or their extensions). Draw three circumcircles (Miquel's circles) to triangles AB´C´, A´BC´, and A´B´C. Miquel's theorem states that these circles intersect in a single point M, called the Miquel point.
Here is a proof using properties of concyclic points to show that the three lines RC, BQ, AP in Fig 2 all intersect at the point F and cut one another at angles of 60°. The triangles RAC, BAQ are congruent because the second is a 60° rotation of the first about A. Hence ∠ARF = ∠ABF and ∠AQF = ∠ACF.
By a symmetric argument, the points AC ∩ ac and BC ∩ bc also exist and belong to the planes of both triangles. Since these two planes intersect in more than one point, their intersection is a line that contains all three points. This proves Desargues's theorem if the two triangles are not contained in the same plane.
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