Proof: Third Angle Theorem
Let's prove the following theorem:
if m∠ABC = m∠XYZ and m∠BCA = m∠YZX, then m∠CAB = m∠ZXY
Proof:
Proof Table
# | Claim | Reason |
---|---|---|
1 | ((m∠ABC) + (m∠BCA)) + (m∠CAB) = 180 | ((m∠ABC) + (m∠BCA)) + (m∠CAB) = 180 |
2 | (m∠ABC) + (m∠BCA) = (m∠XYZ) + (m∠YZX) | if m∠ABC = m∠XYZ and m∠BCA = m∠YZX, then (m∠ABC) + (m∠BCA) = (m∠XYZ) + (m∠YZX) |
3 | ((m∠XYZ) + (m∠YZX)) + (m∠CAB) = 180 | if ((m∠ABC) + (m∠BCA)) + (m∠CAB) = 180 and (m∠ABC) + (m∠BCA) = (m∠XYZ) + (m∠YZX), then ((m∠XYZ) + (m∠YZX)) + (m∠CAB) = 180 |
4 | m∠CAB = 180 + (((m∠XYZ) + (m∠YZX)) ⋅ (-1)) | if ((m∠XYZ) + (m∠YZX)) + (m∠CAB) = 180, then m∠CAB = 180 + (((m∠XYZ) + (m∠YZX)) ⋅ (-1)) |
5 | ((m∠XYZ) + (m∠YZX)) + (m∠ZXY) = 180 | ((m∠XYZ) + (m∠YZX)) + (m∠ZXY) = 180 |
6 | m∠ZXY = 180 + (((m∠XYZ) + (m∠YZX)) ⋅ (-1)) | if ((m∠XYZ) + (m∠YZX)) + (m∠ZXY) = 180, then m∠ZXY = 180 + (((m∠XYZ) + (m∠YZX)) ⋅ (-1)) |
7 | m∠CAB = m∠ZXY | if m∠CAB = 180 + (((m∠XYZ) + (m∠YZX)) ⋅ (-1)) and m∠ZXY = 180 + (((m∠XYZ) + (m∠YZX)) ⋅ (-1)), then m∠CAB = m∠ZXY |
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