Question

Diagonals AC and BD form right angles at point M in parallelogram ABCD. Prove ABCD is a rhombus.

Parallelogram A B C D is shown. Diagonals are drawn from point A to point C and from point B to point D and intersect at point M. Sides A B and D C are parallel and sides B C and A D are parallel. A square is drawn around point M.


Statements

Reasons
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

Answer 1

6. ΔAMB ≅ ΔCMB ≅ ΔCMD ≅ ΔDMA — (AM = MC, MB = MD, and ∠AMB = ∠BMC = ∠CMD = ∠DMA, so each adjacent pair of triangles is congruent by SAS)

Explanation: For example, ΔAMB and ΔCMB have AM = MC (diagonals bisect), MB common, and ∠AMB = ∠BMC (right angles), so ΔAMB ≅ ΔCMB by SAS; similarly for the other adjacent pairs. By CPCTC, AB = BC = CD = DA, hence ABCD is a rhombus.