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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