Proof: Similar Corresponding Bisectors
Let's prove the following theorem:
if △ABC ∼ △XYZ and ray BP bisects ∠ABC and ray YQ bisects ∠XYZ and m∠APC = 180 and m∠XQZ = 180, then (distance BC) / (distance YZ) = (distance BP) / (distance YQ)
Proof:
Proof Table
# | Claim | Reason |
---|---|---|
1 | m∠ABC = m∠XYZ | if △ABC ∼ △XYZ, then m∠ABC = m∠XYZ |
2 | m∠PBC = (m∠ABC) / 2 | if ray BP bisects ∠ABC, then m∠PBC = (m∠ABC) / 2 |
3 | m∠QYZ = (m∠XYZ) / 2 | if ray YQ bisects ∠XYZ, then m∠QYZ = (m∠XYZ) / 2 |
4 | m∠QYZ = (m∠ABC) / 2 | if m∠QYZ = (m∠XYZ) / 2 and m∠ABC = m∠XYZ, then m∠QYZ = (m∠ABC) / 2 |
5 | m∠PBC = m∠QYZ | if m∠PBC = (m∠ABC) / 2 and m∠QYZ = (m∠ABC) / 2, then m∠PBC = m∠QYZ |
6 | m∠BCA = m∠YZX | if △ABC ∼ △XYZ, then m∠BCA = m∠YZX |
7 | m∠BCP = m∠YZQ | if m∠APC = 180 and m∠XQZ = 180 and m∠BCA = m∠YZX, then m∠BCP = m∠YZQ |
8 | △BCP ∼ △YZQ | if m∠PBC = m∠QYZ and m∠BCP = m∠YZQ, then △BCP ∼ △YZQ |
9 | (distance BC) / (distance YZ) = (distance BP) / (distance YQ) | if △BCP ∼ △YZQ, then (distance BC) / (distance YZ) = (distance BP) / (distance YQ) |
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