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
7. ab ≅ bc ≅ cd ≅ da 7. cpctc
8. 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 \(ABCD\) is a parallelogram, \(\angle AMB=\angle BMC = \angle CMD=\angle DMA = 90^{\circ}\), \(AM = MC\), \(MB=MD\)

Step2: Apply SAS congruency theorem

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

• \(AM = MC\) (from statement 5)
• \(\angle AMB=\angle BMC\) (from statement 3)
• \(MB = MB\) (common side)

So, \(\triangle AMB\cong\triangle CMB\) (by SAS)

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

• \(MC = MC\) (common side)
• \(\angle BMC=\angle CMD\) (from statement 3)
• \(MB = MD\) (from statement 5)

So, \(\triangle CMB\cong\triangle CMD\) (by SAS)

For \(\triangle CMD\) and \(\triangle AMD\):

• \(MD = MD\) (common side)
• \(\angle CMD=\angle DMA\) (from statement 3)
• \(MC = AM\) (from statement 5)

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

Answer:

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