A parallelogram is characterized as a quadrilateral whereby the 2 opposite sides room parallel. Among the properties of parallelograms is the the opposite angles are congruent, as we will currently show.
Since this a home of any parallelogram, the is also true of any special parallelogram choose a rectangle, a square, or a rhombus,
ABCD is a parallelogram, AD||BC and AB||DC. Prove that ∠BAD ≅ ∠DCB and also that ∠ADC ≅ ∠CBA
Now, because AB||CD, ∠TAB ≅ ∠ADC, as equivalent angles the parallel lines. And as AD||BC, ∠TAB ≅ ∠ABC, as alternate interior angles. For this reason by the transitive building of equality, ∠ADC ≅ ∠ABC.
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Similarly, due to the fact that BC||AD, ∠PBC ≅ ∠BAD, as matching angles the parallel lines.
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And also as AB||CD, ∠PBC ≅ ∠BCD, as alternate interior angles. So by the transitive home of equality, ∠BAD ≅ ∠BCD.
This is the shortest proof- yet requires reasoning "Outside the Box", or in this case, external the parallelogram.
(1) ABCD is a parallel //Given(2) abdominal muscle || CD //From the an interpretation of a parallelogram(3) ∠TAB ≅ ∠ADC //Corresponding angle of parallel lines(4) ad || BC //From the an interpretation of a parallelogram(5) ∠TAB ≅ ∠ABC //Alternate inner angles(6) ∠ADC ≅ ∠ABC //(3) , (5) , Transitive property of equality(7) BC || advertisement //From the definition of a parallelogram(8) ∠PBC ≅ ∠BAD //Corresponding angle of parallel lines(9) abdominal || CD //From the definition of a parallelogram(10) ∠PBC ≅ ∠BCD //Alternate internal angles(11) ∠BAD ≅ ∠BCD //(8) , (10) , Transitive residential property of equality