Question

1. figure efgh is a parallelogram, and angle hef is a right angle. select all statements that must be true.

a. figure efgh is a rectangle.
b. triangle hef is congruent to triangle gfh.
c. triangle hef is congruent to triangle fgh.
d. segment ed is congruent to segments hd, dg, and df.
e. triangle edh is congruent to triangle hdg.

Explanation:

Step1: Analyze Option A

A parallelogram with one right - angle is a rectangle (by the definition of a rectangle: a parallelogram with four right - angles. Since \(EFHG\) is a parallelogram and \(\angle HEF = 90^{\circ}\), adjacent angles in a parallelogram are supplementary, so all angles are \(90^{\circ}\)).

Step2: Analyze Option B

In \(\triangle HEF\) and \(\triangle GFH\):
\(EF = HG\) (opposite sides of a parallelogram), \(HE=FG\) (opposite sides of a parallelogram), \(HF = FH\) (common side). By SSS (Side - Side - Side) congruence criterion, \(\triangle HEF\cong\triangle GFH\).

Step3: Analyze Option C

In \(\triangle HEF\) and \(\triangle FGH\):
\(HE = FG\) (opposite sides of a parallelogram), \(EF = HG\) (opposite sides of a parallelogram), \(\angle HEF=\angle FGH = 90^{\circ}\) (in a rectangle). By SAS (Side - Angle - Side) congruence criterion, \(\triangle HEF\cong\triangle FGH\).

Step4: Analyze Option D

In a rectangle, the diagonals are equal and bisect each other. So \(EHFG\) is a rectangle (from Option A), diagonals \(EG\) and \(HF\) intersect at \(D\). Then \(ED = HD=DG = DF\) (diagonals of a rectangle bisect each other and are equal).

Step5: Analyze Option E

In \(\triangle EDH\) and \(\triangle HDG\):
\(ED = DG\) (from Option D), \(HD = HD\) (common side), but \(\angle EDH
eq\angle HDG\) (unless it is a square, but we only know it's a rectangle). So \(\triangle EDH\) and \(\triangle HDG\) are not congruent.

Answer:

A. Figure \(EFGH\) is a rectangle, B. Triangle \(HEF\) is congruent to triangle \(GFH\), C. Triangle \(HEF\) is congruent to triangle \(FGH\), D. Segment \(ED\) is congruent to segments \(HD\), \(DG\), and \(DF\)