Search results
Results from the WOW.Com Content Network
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. Two circles are congruent if they have the same diameter.
The orange and green quadrilaterals are congruent; the blue one is not congruent to them. Congruence between the orange and green ones is established in that side BC corresponds to (in this case of congruence, equals in length) JK, CD corresponds to KL, DA corresponds to LI, and AB corresponds to IJ, while angle ∠C corresponds to (equals) angle ∠K, ∠D corresponds to ∠L, ∠A ...
Any two pairs of angles are congruent, [4] which in Euclidean geometry implies that all three angles are congruent: [a] If ∠BAC is equal in measure to ∠B'A'C', and ∠ABC is equal in measure to ∠A'B'C', then this implies that ∠ACB is equal in measure to ∠A'C'B' and the triangles are similar. All the corresponding sides are ...
Given two congruent triangles, all pairs of corresponding interior angles are equal in measure, and all pairs of corresponding sides have the same length. This is a total of six equalities, but three are often sufficient to prove congruence. [42] Some individually necessary and sufficient conditions for a pair of triangles to be congruent are: [43]
Perpendicular bisector of a line segment. The perpendicular bisector of a line segment is a line which meets the segment at its midpoint perpendicularly.; The perpendicular bisector of a line segment also has the property that each of its points is equidistant from segment AB's endpoints:
The two bimedians of a quadrilateral (segments joining midpoints of opposite sides) and the line segment joining the midpoints of the diagonals are concurrent and are all bisected by their point of intersection. [3]: p.125 In a tangential quadrilateral, the four angle bisectors concur at the center of the incircle. [4]
Let AB and BC be two segments of a line a which have no points in common aside from the point B, and, furthermore, let A′B′ and B′C′ be two segments of the same or of another line a′ having, likewise, no point other than B′ in common. Then, if AB ≅ A′B′ and BC ≅ B′C′, we have AC ≅ A′C′.
If A, B are two points on a line a, and if A′ is a point upon the same or another line a′, then, upon a given side of A′ on the straight line a′, we can always find a point B′ so that the segment AB is congruent to the segment A′B′.