Question

1. abcd is a parallelogram 1. given
2. ∠amb, ∠bmc, ∠cmd, and ∠dma are right angles 2. given
3. ∠amb ≅ ∠bmc ≅ ∠cmd ≅ ∠dma 3. right angles are congruent
4. ac bisects bd; bd bisects ac; 4. diagonals of a parallelogram bisect each other
5. am ≅ mc, mb ≅ md 5. definition of a bisector

6.? 6. sas congruency theorem

1. ab ≅ bc ≅ cd ≅ da 7. cpctc
2. figure abcd is a rhombus 8. definition of a rhombus

△abc ≅ △adc ≅ △bad ≅ △bcd
△amb ≅ △cmb ≅ △cmd ≅ △amd
△abc ≅ △adc ≅ △amd ≅ △bmc
△amb ≅ △cmd ≅ △bad ≅ △bcd

Explanation:

Step1: Analyze the given information

We know that \(ABCD\) is a parallelogram, \(\angle AMB,\angle BMC,\angle CMD,\angle DMA\) are right - angles, \(\overline{AM}\cong\overline{MC}\), \(\overline{MB}\cong\overline{MD}\), and we need to use the SAS (Side - Angle - Side) congruency theorem.

Step2: Check the triangles for SAS

For \(\triangle AMB\) and \(\triangle CMB\):

• \(AM = CM\) (from statement 5: \(\overline{AM}\cong\overline{MC}\))
• \(\angle AMB=\angle BMC\) (from statement 3: right - angles are congruent)
• \(BM = BM\) (common side)

So, \(\triangle AMB\cong\triangle CMB\) by SAS.
For \(\triangle CMB\) and \(\triangle CMD\):

• \(CM = CM\) (common side)
• \(\angle BMC=\angle CMD\) (right - angles are congruent)
• \(BM = DM\) (from statement 5: \(\overline{MB}\cong\overline{MD}\))

So, \(\triangle CMB\cong\triangle CMD\) by SAS.
For \(\triangle CMD\) and \(\triangle AMD\):

• \(CM = AM\) (from statement 5: \(\overline{AM}\cong\overline{MC}\))
• \(\angle CMD=\angle AMD\) (right - angles are congruent)
• \(DM = DM\) (common side)

So, \(\triangle CMD\cong\triangle AMD\) by SAS.

Answer:

\(\triangle AMB\cong\triangle CMB\cong\triangle CMD\cong\triangle AMD\)