Mathematics, 03.04.2021 05:30, oroman4650

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Given: DEFG is a parallelogram.

Parallelogram D E F G with diagonal D F. Angle G D F is labeled 1, angle F D E is labeled 2, angle E F D is labeled 3, and angle D F G is labeled 4.

Prove: DE¯¯¯¯¯≅GF¯¯¯¯¯ and DG¯¯¯¯¯¯≅EF¯¯¯¯¯ .

Opposite sides of a parallelogram are parallel, so _ because they are alternate interior angles. By the _, DF¯¯¯¯¯≅DF¯¯¯¯¯ . Therefore, _ by the ASA congruence postulate. That means DE¯¯¯¯¯≅GF¯¯¯¯¯ and DG¯¯¯¯¯¯≅EF¯¯¯¯¯ by CPCTC .

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Given: DEFG is

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