The proof you presented defines properties of a parallelogram and establishes that parallelogram ABCD is specifically a rectangle by showing that it has right angles. Here’s a concise breakdown of each statement and justification:
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Given - ABCD is a parallelogram.
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Given - \( \overline{DB} \cong \overline{AC} \) (The diagonals of a parallelogram bisect each other).
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Opposite sides of a parallelogram are congruent. - \( \overline{DA} \cong \overline{CB} \).
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Reflexive Property of Congruence - \( \overline{AB} \cong \overline{BA} \).
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SSS Congruence Theorem - \( \triangle BAD \cong \triangle ABC \) (All sides of both triangles are shown to be congruent).
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Congruent angles - \( \angle BAD \cong \angle ABC \).
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Definition of congruence - \( m\angle BAD = m\angle ABC \).
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Consecutive angles of a parallelogram are supplementary. - \( m\angle BAD + m\angle ABC = 180^\circ \).
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Same angle added to itself - \( m\angle BAD + m\angle BAD = 180^\circ \).
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Combine like terms - \( 2m\angle BAD = 180^\circ \).
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Division Property of Equality - \( m\angle BAD = 90^\circ \).
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Transitive Property of Equality - \( m\angle ABC = 90^\circ \).
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Opposite angles of a parallelogram are congruent - \( m\angle ADC = m\angle DAB = 90^\circ \).
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Definition of right angles - \( \angle BAD, \angle DCB, \angle ABC, \) and \( \angle ADC \) are all right angles.
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Definition of a rectangle - Since all angles are right angles, ABCD is a rectangle.
This proof successfully demonstrates that parallelogram ABCD is indeed a rectangle by confirming that two pairs of its opposite angles are right angles, and using the important properties of parallelograms related to congruence and supplementary angles.
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