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:

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Given

1	△ABC ∼ △XYZ
2	ray BP bisects ∠ABC
3	ray YQ bisects ∠XYZ
4	m∠APC = 180
5	m∠XQZ = 180

Proof Table

#	Claim	Reason
1	m∠ABC = m∠XYZ	if △ABC ∼ △XYZ, then m∠ABC = m∠XYZ (Similar Angles Property)
2	m∠PBC = (m∠ABC) / 2	if ray BP bisects ∠ABC, then m∠PBC = (m∠ABC) / 2 (Bisector Angle)
3	m∠QYZ = (m∠XYZ) / 2	if ray YQ bisects ∠XYZ, then m∠QYZ = (m∠XYZ) / 2 (Bisector Angle)
4	m∠QYZ = (m∠ABC) / 2	if m∠QYZ = (m∠XYZ) / 2 and m∠ABC = m∠XYZ, then m∠QYZ = (m∠ABC) / 2 (Divide New B)
5	m∠PBC = m∠QYZ	if m∠PBC = (m∠ABC) / 2 and m∠QYZ = (m∠ABC) / 2, then m∠PBC = m∠QYZ (Transitive Property of Equality Variation 1)
6	m∠BCA = m∠YZX	if △ABC ∼ △XYZ, then m∠BCA = m∠YZX (Similar Triangles Property)
7	m∠BCP = m∠YZQ	if m∠APC = 180 and m∠XQZ = 180 and m∠BCA = m∠YZX, then m∠BCP = m∠YZQ (Two Collinear Angles)
8	△BCP ∼ △YZQ	if m∠PBC = m∠QYZ and m∠BCP = m∠YZQ, then △BCP ∼ △YZQ (Similar Triangles Theorem)
9	(distance BC) / (distance YZ) = (distance BP) / (distance YQ)	if △BCP ∼ △YZQ, then (distance BC) / (distance YZ) = (distance BP) / (distance YQ) (Similar Distances)

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