Proof: Rhombus Diagonal Equilateral Triangles
Let's prove the following theorem:
if WXYZ is a rhombus and m∠WXY = 60, then △YZW is an equilateral triangle
Proof:
Given
| 1 | WXYZ is a rhombus |
|---|---|
| 2 | m∠WXY = 60 |
| # | Claim | Reason |
|---|---|---|
| 1 | distance WX = distance XY | if WXYZ is a rhombus, then distance WX = distance XY |
| 2 | distance WX = distance YX | if distance WX = distance XY, then distance WX = distance YX |
| 3 | m∠XWY = m∠XYW | if distance WX = distance YX, then m∠XWY = m∠XYW |
| 4 | m∠YWX = m∠XYW | if m∠XWY = m∠XYW, then m∠YWX = m∠XYW |
| 5 | ((m∠WXY) + (m∠XYW)) + (m∠YWX) = 180 | ((m∠WXY) + (m∠XYW)) + (m∠YWX) = 180 |
| 6 | ((m∠WXY) + (m∠XYW)) + (m∠XYW) = 180 | if ((m∠WXY) + (m∠XYW)) + (m∠YWX) = 180 and m∠YWX = m∠XYW, then ((m∠WXY) + (m∠XYW)) + (m∠XYW) = 180 |
| 7 | (m∠WXY) + ((m∠XYW) ⋅ 2) = 180 | if ((m∠WXY) + (m∠XYW)) + (m∠XYW) = 180, then (m∠WXY) + ((m∠XYW) ⋅ 2) = 180 |
| 8 | 60 + ((m∠XYW) ⋅ 2) = 180 | if (m∠WXY) + ((m∠XYW) ⋅ 2) = 180 and m∠WXY = 60, then 60 + ((m∠XYW) ⋅ 2) = 180 |
| 9 | m∠XYW = 60 | if 60 + ((m∠XYW) ⋅ 2) = 180, then m∠XYW = 60 |
| 10 | m∠WXY = m∠XYW | if m∠WXY = 60 and m∠XYW = 60, then m∠WXY = m∠XYW |
| 11 | m∠XYW = m∠YWX | if m∠YWX = m∠XYW, then m∠XYW = m∠YWX |
| 12 | △WXY is an equilateral triangle | if m∠WXY = m∠XYW and m∠XYW = m∠YWX, then △WXY is an equilateral triangle |
| 13 | distance ZW = distance WX | if WXYZ is a rhombus, then distance ZW = distance WX |
| 14 | distance YZ = distance ZW | if WXYZ is a rhombus, then distance YZ = distance ZW |
| 15 | distance YW = distance WX | if △WXY is an equilateral triangle, then distance YW = distance WX |
| 16 | distance ZW = distance YW | if distance ZW = distance WX and distance YW = distance WX, then distance ZW = distance YW |
| 17 | distance ZW = distance WY | if distance ZW = distance YW, then distance ZW = distance WY |
| 18 | △YZW is an equilateral triangle | if distance YZ = distance ZW and distance ZW = distance WY, then △YZW is an equilateral triangle |
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