Proof: Supplementary Angles 3
Let's prove the following theorem:
if ∠AXB and ∠BXC are supplementary and ∠DXC and ∠BXC are supplementary, then m∠AXB = m∠DXC
Proof:
Given
1 | ∠AXB and ∠BXC are supplementary |
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2 | ∠DXC and ∠BXC are supplementary |
# | Claim | Reason |
---|---|---|
1 | (m∠AXB) + (m∠BXC) = 180 | if ∠AXB and ∠BXC are supplementary, then (m∠AXB) + (m∠BXC) = 180 |
2 | (m∠DXC) + (m∠BXC) = 180 | if ∠DXC and ∠BXC are supplementary, then (m∠DXC) + (m∠BXC) = 180 |
3 | m∠AXB = 180 + ((m∠BXC) ⋅ (-1)) | if (m∠AXB) + (m∠BXC) = 180, then m∠AXB = 180 + ((m∠BXC) ⋅ (-1)) |
4 | m∠DXC = 180 + ((m∠BXC) ⋅ (-1)) | if (m∠DXC) + (m∠BXC) = 180, then m∠DXC = 180 + ((m∠BXC) ⋅ (-1)) |
5 | m∠AXB = m∠DXC | if m∠AXB = 180 + ((m∠BXC) ⋅ (-1)) and m∠DXC = 180 + ((m∠BXC) ⋅ (-1)), then m∠AXB = m∠DXC |
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