QUESTION IMAGE
Question
proving when a parallelogram is a rectangle
given: wxyz is a parallelogram.
(overline{zx}congoverline{wy})
prove: wxyz is a rectangle
statements
- wxyz is a parallelogram
- (overline{yx}congoverline{yx})
- (overline{zy}congoverline{wx})
- (mangle zyx + mangle wxy = 180)
- (overline{zx}congoverline{wy})
- (\triangle zyxcong\triangle wxy)
reasons
- given
- reflexive property
- opposite sides of a parallelogram are congruent
- consecutive angles of a parallelogram are supplementary
- given
- sss
Step1: Use congruent triangles property
Since \(\triangle ZYX\cong\triangle WXY\) (by SSS as \(ZY = WX\), \(YX=YX\), \(ZX = WY\)), then \(\angle ZYX\cong\angle WXY\) (corresponding parts of congruent triangles are congruent). So \(m\angle ZYX=m\angle WXY\)
Step2: Use angle - sum property of parallelogram
In parallelogram \(WXYZ\), \(\angle ZYX\) and \(\angle WXY\) are consecutive angles. We know that \(m\angle ZYX + m\angle WXY=180^{\circ}\) (consecutive angles of a parallelogram are supplementary). Substitute \(m\angle WXY\) with \(m\angle ZYX\) (from step 1), we get \(m\angle ZYX+m\angle ZYX = 180^{\circ}\), i.e., \(2(m\angle ZYX)=180^{\circ}\)
Step3: Solve for the angle measure
Divide both sides of \(2(m\angle ZYX)=180^{\circ}\) by 2. We have \(m\angle ZYX=\frac{180^{\circ}}{2}=90^{\circ}\)
Step4: Use the definition of a rectangle
A parallelogram with one right angle (\(m\angle ZYX = 90^{\circ}\)) is a rectangle. So \(WXYZ\) is a rectangle
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The statements in order: \(m\angle ZYX=m\angle WXY\); \(m\angle ZYX + m\angle ZYX=180\); \(\angle ZYX\cong\angle WXY\); \(2(m\angle ZYX)=180\); \(m\angle ZYX = 90\); \(WXYZ\) is a rectangle