Proof: Concurrent Angle Bisectors

Let's prove the following theorem:

if ray AS bisects ∠CAB and ray BT bisects ∠ABC and m∠APS = 180 and m∠BPT = 180 and PZ ⊥ ZA and PX ⊥ XA and PX ⊥ XB and PY ⊥ YB and PZ ⊥ ZC and PY ⊥ YC, then ray CP bisects ∠BCA

A C B S T P X Y Z

Proof:

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Given
1 ray AS bisects ∠CAB
2 ray BT bisects ∠ABC
3 m∠APS = 180
4 m∠BPT = 180
5 PZ ⊥ ZA
6 PX ⊥ XA
7 PX ⊥ XB
8 PY ⊥ YB
9 PZ ⊥ ZC
10 PY ⊥ YC
Proof Table
# Claim Reason
1 distance PX = distance PZ if ray AS bisects ∠CAB and m∠APS = 180 and PX ⊥ XA and PZ ⊥ ZA, then distance PX = distance PZ
2 distance PX = distance PY if ray BT bisects ∠ABC and m∠BPT = 180 and PX ⊥ XB and PY ⊥ YB, then distance PX = distance PY
3 distance PZ = distance PY if distance PX = distance PZ and distance PX = distance PY, then distance PZ = distance PY
4 ray CP bisects ∠BCA if distance PZ = distance PY and PZ ⊥ ZC and PY ⊥ YC, then ray CP bisects ∠BCA

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