Proof: Complementary Angles
Let's prove the following theorem:
if ∠AXB and ∠BXC are complementary and ∠BXC and ∠CXD are complementary, then m∠AXB = m∠CXD
Proof:
Given
1 | ∠AXB and ∠BXC are complementary |
---|---|
2 | ∠BXC and ∠CXD are complementary |
# | Claim | Reason |
---|---|---|
1 | (m∠AXB) + (m∠BXC) = 90 | if ∠AXB and ∠BXC are complementary, then (m∠AXB) + (m∠BXC) = 90 |
2 | (m∠BXC) + (m∠CXD) = 90 | if ∠BXC and ∠CXD are complementary, then (m∠BXC) + (m∠CXD) = 90 |
3 | m∠AXB = 90 + ((m∠BXC) ⋅ (-1)) | if (m∠AXB) + (m∠BXC) = 90, then m∠AXB = 90 + ((m∠BXC) ⋅ (-1)) |
4 | m∠CXD = 90 + ((m∠BXC) ⋅ (-1)) | if (m∠BXC) + (m∠CXD) = 90, then m∠CXD = 90 + ((m∠BXC) ⋅ (-1)) |
5 | m∠AXB = m∠CXD | if m∠AXB = 90 + ((m∠BXC) ⋅ (-1)) and m∠CXD = 90 + ((m∠BXC) ⋅ (-1)), then m∠AXB = m∠CXD |
Comments
Please log in to add comments