Proof: Similar Corresponding Altitudes
Let's prove the following theorem:
if △ABC ∼ △XYZ and ∠BPC is a right angle and ∠YQZ is a right angle and m∠APC = 180 and m∠XQZ = 180, then (distance BP) / (distance YQ) = (distance BC) / (distance YZ)
Proof:
Proof Table
# | Claim | Reason |
---|---|---|
1 | m∠BCA = m∠YZX | if △ABC ∼ △XYZ, then m∠BCA = m∠YZX |
2 | m∠BCP = m∠YZQ | if m∠APC = 180 and m∠XQZ = 180 and m∠BCA = m∠YZX, then m∠BCP = m∠YZQ |
3 | m∠BPC = 90 | if ∠BPC is a right angle, then m∠BPC = 90 |
4 | m∠YQZ = 90 | if ∠YQZ is a right angle, then m∠YQZ = 90 |
5 | m∠BPC = m∠YQZ | if m∠BPC = 90 and m∠YQZ = 90, then m∠BPC = m∠YQZ |
6 | m∠CPB = m∠ZQY | if m∠BPC = m∠YQZ, then m∠CPB = m∠ZQY |
7 | △CPB ∼ △ZQY | if m∠BCP = m∠YZQ and m∠CPB = m∠ZQY, then △CPB ∼ △ZQY |
8 | (distance BP) / (distance YQ) = (distance BC) / (distance YZ) | if △CPB ∼ △ZQY, then (distance BP) / (distance YQ) = (distance BC) / (distance YZ) |
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