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
Proof:
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 |
# | 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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