given: quadrilateral fghi,
$ \overline { g f } \parallel \overline { h k } $, and
$ \overline { g f } \cong \overline { h k } $
prove: fghi is a trapezoid.
1. quadrilateral fghi,
$ \overline { g f } \parallel \overline { h k } $, and $ \overline { g f } \cong \overline { h k } $
2.
3. $ \overline { g h } \parallel \overline { f k } $ and $ \overline { g h } \parallel \overline { f i } $
1. given
2.
3.

Question

Accepted Answer

# Explanation:
## Step1: Prove that quadrilateral $FGHK$ is a parallelogram
Since $GF\parallel HK$ and $GF\cong HK$, by the definition of a parallelogram (a quadrilateral with one pair of opposite sides parallel and congruent is a parallelogram), quadrilateral $FGHK$ is a parallelogram.
## Step2: Use the property of a parallelogram
In parallelogram $FGHK$, $GH\parallel FK$ (opposite sides of a parallelogram are parallel).
## Step3: Use the transitive property of parallel lines
Since $FK\parallel FI$ (co - linear points $K$ and $I$ on line $FK$ and $FI$), and $GH\parallel FK$, then $GH\parallel FI$ (if $a\parallel b$ and $b\parallel c$, then $a\parallel c$).
## Step4: Use the definition of a trapezoid
A trapezoid is a quadrilateral with at least one pair of parallel sides. In quadrilateral $FGHI$, $GH\parallel FI$, so $FGHI$ is a trapezoid.

# Answer:
2. Quadrilateral $FGHK$ is a parallelogram; Reason: A quadrilateral with one pair of opposite sides parallel and congruent is a parallelogram.
3. Reason: Transitive property of parallel lines.

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Question

Explanation:

Step1: Prove that quadrilateral \(FGHK\) is a parallelogram

Step2: Use the property of a parallelogram

Step3: Use the transitive property of parallel lines

Step4: Use the definition of a trapezoid

Answer: