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:

View as a tree | View dependent proofs | Try proving it

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 (Supplementary Angles Property)
2	(m∠DXC) + (m∠BXC) = 180	if ∠DXC and ∠BXC are supplementary, then (m∠DXC) + (m∠BXC) = 180 (Supplementary Angles Property)
3	m∠AXB = 180 + ((m∠BXC) ⋅ (-1))	if (m∠AXB) + (m∠BXC) = 180, then m∠AXB = 180 + ((m∠BXC) ⋅ (-1)) (Add Term to Both Sides 6)
4	m∠DXC = 180 + ((m∠BXC) ⋅ (-1))	if (m∠DXC) + (m∠BXC) = 180, then m∠DXC = 180 + ((m∠BXC) ⋅ (-1)) (Add Term to Both Sides 6)
5	m∠AXB = m∠DXC	if m∠AXB = 180 + ((m∠BXC) ⋅ (-1)) and m∠DXC = 180 + ((m∠BXC) ⋅ (-1)), then m∠AXB = m∠DXC (Transitive Property of Equality Variation 1)

Comments

Please log in to add comments