Question

given: jkl.m is a rhombus;
overline {kp}cong overline {jq};overline {lp}cong overline {kq}
prove: jkl.m is a square.

1. angle k p cong angle m
2. ( m angle m = 90 ), ( m angle k l m = 90 )
3. jkl.m is a square.

Explanation:

Step1: Recall the properties of a square

A square is a rhombus with one right - angle.

Step2: Analyze the given information

We are given that \(JKLM\) is a rhombus. In a rhombus, if one angle is \(90^{\circ}\), then all angles are \(90^{\circ}\) (because consecutive angles of a parallelogram (a rhombus is a parallelogram) are supplementary, and if \(\angle M = 90^{\circ}\), then \(\angle KLM=90^{\circ}\), \(\angle J = 90^{\circ}\) and \(\angle K = 90^{\circ}\) using the property that pairs of consecutive angles in a rhombus/parallelogram are supplementary).

Step3: Use the definition of a square

Since \(JKLM\) is a rhombus (all sides are equal) and has four right - angles (\(m\angle M=m\angle KLM = 90^{\circ}\), and other angles can be shown to be \(90^{\circ}\) as well), by the definition of a square (a quadrilateral that is a rhombus and a rectangle (a quadrilateral with four right - angles)), \(JKLM\) is a square.

Answer:

A rhombus with one right - angle is a square. Since \(JKLM\) is a rhombus and \(m\angle M = 90^{\circ}\), \(JKLM\) is a square.