Quiz (1 point)
Prove that:
The following properties may be helpful:
- distance AB = distance BA
if the following are true:
- a = 90
- b = 90
then a + b = 180
- if (m∠ABC) + (m∠DEF) = 180, then ∠ABC and ∠DEF are supplementary
- if ∠WST and ∠YTS are supplementary, then WS || YT
- if quadrilateral ABCD is a trapezoid, then AB || DC
- if (AB || CD) and (m∠AXY = 180) and (m∠XYB = 180), then XY || CD
- if AB || CD, then CD || AB
- if (AB || CD) and (AC || BD), then ABDC is a parallelogram
- if WXYZ is a parallelogram, then distance WX = distance ZY
- if distance AB = distance CD, then distance BA = distance DC
- if m∠ABC = 180, then m∠XAB = m∠XAC
- if (m∠DXY = 90) and (m∠DXY = m∠DXB), then m∠BXD = 90
- if m∠ABC = 90, then ∠ABC is a right angle
- if m∠ABC = 180, then m∠XCB = m∠XCA
- if (m∠DXY = 90) and (m∠DXY = m∠DXB), then m∠BXD = 90
- if m∠ABC = 90, then ∠ABC is a right angle
if a = b, then b = a
- if (∠ABC is a right angle) and (∠XYZ is a right angle) and (distance AC = distance XZ) and (distance BC = distance YZ), then △ABC ≅ △XYZ
- if △ABC ≅ △DEF, then m∠CAB = m∠FDE
- if (m∠ABC = 180) and (m∠BCD = 180), then m∠ABD = 180
- if (m∠ABC = 180) and (m∠BCD = 180), then m∠ACD = 180
- if (m∠DBX = m∠CAY) and (m∠AXB = 180) and (m∠AYB = 180), then m∠DBA = m∠CAB
- if △ABC ≅ △DEF, then distance CA = distance FD
- if (distance AB = distance DE) and (m∠ABC = m∠DEF) and (distance BC = distance EF), then △ABC ≅ △DEF
- if △ABC ≅ △DEF, then distance AC = distance DF
Please write your proof in the table below. Each row should contain one claim. The last claim is the statement that you are trying to prove.