Proof: Angles of an Isosceles Triangle
Let's prove the following theorem:
if distance AX = distance BX, then m∠BAX = m∠ABX
Proof:
Proof Table
# | Claim | Reason |
---|---|---|
1 | distance AM = distance MB | if M is the midpoint of line AB, then distance AM = distance MB |
2 | distance MA = distance MB | if distance AM = distance MB, then distance MA = distance MB |
3 | distance XM = distance XM | distance XM = distance XM |
4 | △AXM ≅ △BXM | if distance AX = distance BX and distance XM = distance XM and distance MA = distance MB, then △AXM ≅ △BXM |
5 | m∠MAX = m∠MBX | if △AXM ≅ △BXM, then m∠MAX = m∠MBX |
6 | m∠AMB = 180 | if M is the midpoint of line AB, then m∠AMB = 180 |
7 | m∠MBX = m∠ABX | if m∠AMB = 180, then m∠MBX = m∠ABX |
8 | m∠MAX = m∠BAX | if m∠AMB = 180, then m∠MAX = m∠BAX |
9 | m∠BAX = m∠ABX | if m∠MAX = m∠MBX and m∠MAX = m∠BAX and m∠MBX = m∠ABX, then m∠BAX = m∠ABX |
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