Asked by Ian
Prove that quadrilateral ABCD is a parallelogram. How do you know that the figure is a parallelogram? A. Adjacent sides of the figure are parallel. B. Adjacent sides of the figure are congruent. C. Adjacent angles of the figure are congruent. D. Opposite sides of the figure are congruent.
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Answer: D. Opposite sides of the figure are congruent.
Proof: Suppose AB = CD and BC = AD. Draw diagonal AC. In triangles ABC and CDA we have
- AB = CD (given),
- BC = AD (given),
- AC = AC (common).
Thus triangles ABC and CDA are congruent by SSS. Corresponding angles are equal, so ∠CAB = ∠DCA and ∠BCA = ∠DAC. These equalities are alternate interior angles made by lines AB and CD and by lines BC and AD with transversal AC, so AB ∥ CD and BC ∥ AD. With both pairs of opposite sides parallel, ABCD is a parallelogram.
(Notes: A is impossible for a nondegenerate quadrilateral, B is satisfied by kites which are not parallelograms, and C can hold in an isosceles trapezoid that is not a parallelogram.)
Proof: Suppose AB = CD and BC = AD. Draw diagonal AC. In triangles ABC and CDA we have
- AB = CD (given),
- BC = AD (given),
- AC = AC (common).
Thus triangles ABC and CDA are congruent by SSS. Corresponding angles are equal, so ∠CAB = ∠DCA and ∠BCA = ∠DAC. These equalities are alternate interior angles made by lines AB and CD and by lines BC and AD with transversal AC, so AB ∥ CD and BC ∥ AD. With both pairs of opposite sides parallel, ABCD is a parallelogram.
(Notes: A is impossible for a nondegenerate quadrilateral, B is satisfied by kites which are not parallelograms, and C can hold in an isosceles trapezoid that is not a parallelogram.)
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