Quiz (1 point)
Prove that:
△ABC ∼ △XYZ
The following properties may be helpful:
- if AB || CD, then CD || AB
- if (WX || YZ) and (m∠WYR = 180), then m∠ZYR = m∠XWY
- if m∠ABC = 180, then m∠XAB = m∠XAC
if the following are true:
- a = b
- b = c
then a = c
if a = b, then b = a
- if (m∠ADC = 180) and (m∠BEC = 180), then m∠ACB = m∠DCE
- if (m∠CAB = m∠ZXY) and (m∠ABC = m∠XYZ), then △ABC ∼ △XYZ
- if △ABC ∼ △DEF, then △ACB ∼ △DFE
- if △ABC ∼ △DEF, then (distance ED) / (distance BA) = (distance DF) / (distance AC)
- if distance AB = distance CD, then distance DC = distance BA
if a = b, then b = a
if the following are true:
- a / b = c
- d = a
then d / b = c
if the following are true:
- a = b
- b = c
then a = c
if the following are true:
- a = b
- b = c
then c = a
if a / c = b / c, then a = b
if a = b, then b = a
- if △ABC ∼ △DEF, then (distance DF) / (distance AC) = (distance FE) / (distance CB)
if the following are true:
- a = b
- b = c
then a = c
if the following are true:
- a = b
- a = c
then b = c
if a / c = b / c, then b = a
- if (distance AB = distance DE) and (distance BC = distance EF) and (distance CA = distance FD), then △ABC ≅ △DEF
- if △ABC ≅ △DEF, then △ABC ∼ △DEF
- if (△ABC ∼ △DEF) and (△XYZ ∼ △DEF), then △ABC ∼ △XYZ
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.