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Question
given: parallelogram lmno; mo ⊥ ln prove: lmno is a rhombus. statements reasons 1. lmno is a parallelogram 1. given 2. mo ⊥ ln 2. given 3. ∠mpn and ∠mpl are right angles 3. def of perpendicular 4. ∠mpn ≅ ∠mpl 4. 5. np ≅ lp 5. diagonals of a parallelogram bisect each other 6. mp ≅ mp 6. 7. △mnp ≅ △mlp 8. sas 8. lm ≅ mn 8. cpctc
Step1: Define right - angle congruence
All right angles are congruent. So, since $\angle MPN$ and $\angle MPL$ are right angles, $\angle MPN\cong\angle MPL$ because all right angles have a measure of $90^{\circ}$ and angles with the same measure are congruent.
Step2: Reflexive property of congruence
$\overline{MP}\cong\overline{MP}$ by the reflexive property of congruence, which states that any segment is congruent to itself.
Step3: Prove triangle congruence
In $\triangle MNP$ and $\triangle MLP$, we have $\angle MPN\cong\angle MPL$ (from step 1), $\overline{NP}\cong\overline{LP}$ (diagonals of a parallelogram bisect each other), and $\overline{MP}\cong\overline{MP}$ (from step 2). By the Side - Angle - Side (SAS) congruence criterion, $\triangle MNP\cong\triangle MLP$.
Step4: Use CPCTC
Since $\triangle MNP\cong\triangle MLP$, by the Corresponding Parts of Congruent Triangles are Congruent (CPCTC) property, $\overline{LM}\cong\overline{MN}$. In a parallelogram, if two adjacent sides are congruent, then the parallelogram is a rhombus.
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The reasons for statements 4 and 6 are:
- All right angles are congruent.
- Reflexive property of congruence.