Proof: Triangle And Line Parallel
Let's prove the following theorem:
if distance WY = distance XY and ray YP bisects ∠XYZ and m∠WYZ = 180, then WX || YP
Proof:
Proof Table
# | Claim | Reason |
---|---|---|
1 | m∠XYP = m∠PYZ | if ray YP bisects ∠XYZ, then m∠XYP = m∠PYZ |
2 | point P lies in interior of ∠XYZ | if ray YP bisects ∠XYZ, then point P lies in interior of ∠XYZ |
3 | (m∠XYP) + (m∠PYZ) = m∠XYZ | if point P lies in interior of ∠XYZ, then (m∠XYP) + (m∠PYZ) = m∠XYZ |
4 | m∠XWY = m∠WXY | if distance WY = distance XY, then m∠XWY = m∠WXY |
5 | (m∠XWY) + (m∠WXY) = m∠XYZ | if m∠WYZ = 180, then (m∠XWY) + (m∠WXY) = m∠XYZ |
6 | (m∠WXY) + (m∠WXY) = m∠XYZ | if (m∠XWY) + (m∠WXY) = m∠XYZ and m∠XWY = m∠WXY, then (m∠WXY) + (m∠WXY) = m∠XYZ |
7 | m∠WXY = (m∠XYZ) ⋅ (1 / 2) | if (m∠WXY) + (m∠WXY) = m∠XYZ, then m∠WXY = (m∠XYZ) ⋅ (1 / 2) |
8 | (m∠XYP) + (m∠XYP) = m∠XYZ | if (m∠XYP) + (m∠PYZ) = m∠XYZ and m∠XYP = m∠PYZ, then (m∠XYP) + (m∠XYP) = m∠XYZ |
9 | m∠XYP = (m∠XYZ) ⋅ (1 / 2) | if (m∠XYP) + (m∠XYP) = m∠XYZ, then m∠XYP = (m∠XYZ) ⋅ (1 / 2) |
10 | m∠WXY = m∠XYP | if m∠WXY = (m∠XYZ) ⋅ (1 / 2) and m∠XYP = (m∠XYZ) ⋅ (1 / 2), then m∠WXY = m∠XYP |
11 | WX || YP | if m∠WXY = m∠XYP, then WX || YP |
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