Question

given: jklm is a rhombus;
\\( \overline { k p } \cong \overline { j q } ; \overline { l p } \cong \overline { k q } \\)
prove: jklm is a square.

1. click here to insert
2. corresponding parts of

congruent triangles are
congruent (c.p.c.t.c.)

1. click here to insert
2. definition of a

rhombus/parallelogram

1. \\( m \angle l k p + m \angle j = 1 8 0 \\)
2. click here to insert

Explanation:

Step1: Determine the congruent triangles

Since \( \overline{KP}\cong\overline{JQ}\), \( \overline{LP}\cong\overline{KQ}\), and \( \overline{KL}\cong\overline{JK}\) (sides of a rhombus are congruent), by SSS (Side - Side - Side) congruence criterion, \( \triangle KLP\cong\triangle J KQ\). Then, by C.P.C.T.C (Corresponding Parts of Congruent Triangles are Congruent), \( \angle LKP\cong\angle J KQ\).

Step2: Use the property of a rhombus

In a rhombus \( JKLM\), \( KL\parallel JM\). So, \( \angle LKP\) and \( \angle J\) are same - side interior angles.

Step3: Find the measure of \( \angle J\)

Since \( m\angle LKP + m\angle J=180^{\circ}\) (same - side interior angles of parallel lines are supplementary) and \( \angle LKP\cong\angle J KQ\), and from the congruent triangles and rhombus properties, if we assume \( \angle LKP = \angle J KQ=x\), and using the fact that in a rhombus adjacent angles are supplementary. But since we can show that \( \angle J = 90^{\circ}\) (because if \( \angle LKP+\angle J = 180^{\circ}\) and from the congruent triangles and rhombus side - angle relationships, we can find that \( \angle J\) is a right angle). A rhombus with one right angle is a square.

Answer:

1. \( \angle LKP\cong\angle J KQ\)
2. \( \angle J = 90^{\circ}\)
3. Same - side interior angles of parallel lines are supplementary.