In a right △ABC, CD (D ∈ AB) is the angle bisector of the right ∠C. Two segments, DF (F ∈ AC) and DE (E ∈BC) are drawn parallel to the legs of the triangle. Prove that DFCE is a square.
m∠CED = 180° − m∠___= by reason
m∠CFD = 180° − m∠_= __by reason

You an see that they gave you parallel lines in DFCE. You can use these to input parallel line contingencies. You can then prove, using Same Side Interior Angles, that angle CED and angle CFD are 90 degrees, making the shape a square as well. Hope this helped :)

In a right △ABC, CD (D ∈ AB) is the angle bisector of the right ∠C. Two segments, DF (F ∈ AC) and DE (E ∈BC) are drawn parallel to the legs of the triangle. Prove that DFCE is a square. m∠CED = 180° − m∠_______= ______by reason m∠CFD = 180° − m∠_______= ______by reason

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In a right △ABC, CD (D ∈ AB) is the angle bisector of the right ∠C. Two segments, DF (F ∈ AC) and DE (E ∈BC) are drawn parallel to the legs of the triangle. Prove that DFCE is a square.
m∠CED = 180° − m∠___= by reason
m∠CFD = 180° − m∠_= __by reason