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A parallelogram is characterized as a quadrilateral where the two opposite sides room parallel. Among the nature of parallelograms is that the the opposite angles space congruent, together we will now show.

Since this a building of any kind of parallelogram, that is also true of any special parallelogram like a rectangle, a square, or a rhombus,

Problem

ABCD is a parallelogram, AD||BC and also AB||DC. Prove the ∠BAD ≅ ∠DCB and also that ∠ADC ≅ ∠CBA

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Now, since AB||CD, ∠TAB ≅ ∠ADC, as equivalent angles of parallel lines. And as AD||BC, ∠TAB ≅ ∠ABC, as alternate interior angles. For this reason by the transitive property of equality, ∠ADC ≅ ∠ABC.

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Similarly, due to the fact that BC||AD, ∠PBC ≅ ∠BAD, as corresponding angles that parallel lines.

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And also as AB||CD, ∠PBC ≅ ∠BCD, as alternative 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, outside the parallelogram.

Proof #3

(1) ABCD is a parallelogram //Given(2) abdominal muscle || CD //From the meaning of a parallelogram(3) ∠TAB ≅ ∠ADC //Corresponding angle of parallel lines(4) ad || BC //From the definition of a parallelogram(5) ∠TAB ≅ ∠ABC //Alternate interior angles(6) ∠ADC ≅ ∠ABC //(3) , (5) , Transitive building of equality(7) BC || ad //From the meaning of a parallelogram(8) ∠PBC ≅ ∠BAD //Corresponding angles of parallel lines(9) ab || CD //From the meaning of a parallelogram(10) ∠PBC ≅ ∠BCD //Alternate internal angles(11) ∠BAD ≅ ∠BCD //(8) , (10) , Transitive residential property of equality