Subtract Both Sides
Add Term to Both Sides 6
Transitive Property of Equality Variation 1
Supplementary Angles 3

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

A B D C X

Proof:

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Given
1 ∠AXB and ∠BXC are supplementary
2 ∠DXC and ∠BXC are supplementary
Proof Table
# 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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