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)

C A B P Z Y X Q

Proof:

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Given
1 △ABC ∼ △XYZ
2 ∠BPC is a right angle
3 ∠YQZ is a right angle
4 m∠APC = 180
5 m∠XQZ = 180
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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