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[6] [7] The convex hull of an antiparallelogram is an isosceles trapezoid, and every antiparallelogram may be formed from an isosceles trapezoid (or its special cases, the rectangles and squares) by replacing two parallel sides by the two diagonals of the trapezoid. [4] Because an antiparallelogram forms two congruent triangular regions of the ...
More specifically, let A, B, C and D be four points on a circle such that the lines AC and BD are perpendicular. Denote the intersection of AC and BD by M. Drop the perpendicular from M to the line BC, calling the intersection E. Let F be the intersection of the line EM and the edge AD. Then, the theorem states that F is the midpoint AD.
Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin( α + β ) = sin α cos β + cos α sin ...
Two pairs of opposite sides are parallel (by definition). Two pairs of opposite sides are equal in length. Two pairs of opposite angles are equal in measure. The diagonals bisect each other. One pair of opposite sides is parallel and equal in length. Adjacent angles are supplementary. Each diagonal divides the quadrilateral into two congruent ...
The diagonals are perpendicular. The two line segments connecting opposite points of tangency have equal lengths. One pair of opposite tangent lengths have equal lengths. The bimedians have equal lengths. The products of opposite sides are equal. The center of the incircle lies on the diagonal that is the axis of symmetry.
A kite is an orthodiagonal quadrilateral in which one diagonal is a line of symmetry.The kites are exactly the orthodiagonal quadrilaterals that contain a circle tangent to all four of their sides; that is, the kites are the tangential orthodiagonal quadrilaterals.
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 ...
The two legs of both triangles have a taxicab length 2, but the hypotenuses are not congruent. This counterexample eliminates AAS, ASA, and SAS. It also eliminates AASS, AAAS, and even ASASA. Having three congruent angles and two sides does not guarantee triangle congruence in taxicab geometry.