Mathematics, 13.10.2020 01:01 robert7248
Drag and drop the answers to the boxes to correctly complete the proof.
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 Response area because they are alternate interior angles. By the reflexive property of congruence, DF¯¯¯¯¯≅DF¯¯¯¯¯ . Therefore, △DGF≅△FED by the Response area . That means DE¯¯¯¯¯≅GF¯¯¯¯¯ and DG¯¯¯¯¯¯≅EF¯¯¯¯¯ by Response area.
Answers: 1
and 
by CPCTC
and  
(Opposite sides of a parallelogram are parallel)
(By the reflexive property of congruence)
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Given: DEFG is a parallelog...
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