Fun Geometry Problem with Solution #100



٨ x = 6°
٨



(1) ∠ACB = 108°
∵ ∠BAC =∠ABC ∆ABC 繠∆˹Ҩ ՠ∠C ʹ AC = BC

(2) BP ͡ѧش D ·∠DAP = 24° ∠CAD = 6°,∠ADP = 126°Р∠APD = 30°
α= 6°
ѧࡵҠ☐ACBD AC = BC,∠A =α,∠B = 2αР∠C = 120°- 2α BC = BD (Click ʹԸվ٨⨷ 2) ∆BCD 繠∆˹Ҩ ՠ∠B (= 12°) ʹ ∠BCD (=∠BDC) = 84° ∠ACD = 24°

(3) CD ͡仾 AP ش E ∠ADE = 30° ∠EDP = 96°Р∠DEP = 54°

(4) ˹ش O 繨شٹҧͧǧՠ∆ADP Ṻ ...
AO = DO = OP
∠AOD = 2(∠APD) ∠AOD = 60°
∠DOP = 2(∠DAP) ∠DOP = 48°
∵ AO = DO Р∠AOD = 60° ∆ADO 繠∆ҹ ...
AD = AO AD = OP
∠DAO = 60° ∠OAP = 36°
∠ADO = 60° ∠EDO = 30°
ѧࡵҠ∆DEO∆ADE ¤ѹẺ -- (DO = AD,∠EDO = ∠ADE, DE = DE) ∠DOE =∠DAE ∠DOE = 24°
∵ AO = OP ∆AOP 繠∆˹Ҩ ՠ∠O ʹ ∠APO =∠OAP ∠APO = 36°
ԨóҠ∆EOP Ҡ∠OEP = 72° ∠OEP =∠EOP ∆EOP 繠∆˹Ҩ ՠ∠P ʹ EP = OP EP = AD

(5) DE = k
˹ش Q 繨شٹҧͧǧՠ∆DEP Ṻ ...
DQ = EQ = PQ
∠DQE = 2(∠DPE) ∠DQE = 60°
∠DQP = 2(∠DEP) ∠DQP = 108°
∵ DQ = EQ Р∠DQE = 60° ∆DEQ 繠∆ҹ EQ = DE EQ = k PQ = k
∵ EQ = PQ ∆EPQ 繠∆˹Ҩ ՠ∠Q (= 168°) ʹ ∠PEQ =∠EPQ = 6°

(6) ˹ش F AC AF = k
Ҡ∆ADF∆EPQ ¤ѹẺ -- (AD = EP,∠DAF =∠PEQ, AF = EQ) DF = PQ DF = k
͡ҡ ѧҠ∠AFD =∠EQP ∠AFD = 168° ∠CFD = 12°

(7) FC ͡ѧش G ∠DGF = 12° ∠DGF =∠DFG ∆DFG 繠∆˹Ҩ ՠ∠D ʹ DG = DF DG = k
͡ҡ ѧҠ∠DCG = 156° ∠CDG = 12°(∠GDP = 72°) ∠CDG =∠CGD ∆CDG 繠∆˹Ҩ ՠ∠C ʹ CD = CG

(8) ѧࡵҠ∆DEG∆EPQ ¤ѹẺ -- (DE = EQ,∠EDG =∠EQP, DG = PQ) ∠DEG =∠PEQ ∠DEG = 6° ∠GEP = 48°
͡ҡ ѧ EG = EP ∆EGP 繠∆˹Ҩ ՠ∠E (= 48°) ʹ ∠EPG (=∠EGP) = 66° ∠DPG = 36°
ԨóҠ∆DGP Ҡ∠DGP = 72° ∠DGP =∠GDP ∆DGP 繠∆˹Ҩ ՠ∠P ʹ DP = GP
ѧࡵҠ∆CDP∆CGP ¤ѹẺ -- (CD = CG, DP = GP, CP = CP) ∠DCP =∠GCP ∠DCP = (∠DCG)/2 = 78° ∠BCP = x = 6° Q.E.D.




Create Date : 15 Ҿѹ 2558
Last Update : 15 Ҿѹ 2558 0:00:00 .
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