Fun Geometry Problem with Solution #155



٨ x= 15°
٨



(1) ԨóҠ∆ABC Ҡ∠CBD = 180°- 5x
ԨóҠ∆BCD Ҡ∠BDC = 3x

(2) ҧǧͺ∆ABD С˹ͺǧѴѺ CD ش P ☐ABPD öṺǧ ∠BAP = ∠BDP, ∠APB =∠ADB Р∠APD =∠ABD ∠BAP = 3x,∠APB = 3x Р∠APD = 2x
∵ ∠BAP =∠APB ∆ABP 繠∆˹Ҩ ՠ∠B ʹ AB = BP
∵ ∠CBP +∠BCP =∠BPD ∠CBP + 2x = 5x ∠CBP = 3x

(3) ˹ش Q AC BQ = AB ∆ABQ 繠∆˹Ҩ ՠ∠B ʹ ∠AQB =∠BAQ ∠AQB = 2x ∠CBQ = x ∠PBQ = 2x
∵ ∠CBQ =∠BCQ ∆BCQ 繠∆˹Ҩ ՠ∠Q ʹ CQ = BQ CQ = AB

(4) ѧࡵҠ☐BPCQ BP = BQ = CQ,∠B = 2x Р∠C = x ∠BPC = 120°- x (Click ʹԸվ٨⨷ 1)

(5) ԨóҠ∆BCP Ҡ∠CBP +∠BCP +∠BPC = 180° 3x + 2x + (120°- x) = 180° x = 15° Q.E.D.




Create Date : 30 áҤ 2558
Last Update : 30 áҤ 2558 0:00:00 .
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Fun Geometry Problem with Solution #154



٨ x= 10°
٨



(1)∠ACP = 20°- x Р∠BPC = 50°

(2) ˹ش Q Ҿз͹ͧش P ҹ BC ∆BCQ∆BCP BQ = BP,∠CBQ =∠CBP = 30°Р∠BQC = ∠BPC = 50°
∵ BP = BQ Р∠PBQ = 60° ∆BPQ 繠∆ҹ BP = PQ
͡ҡ ѧҠ∠BPQ =∠BQP = 60° ∠CPQ =∠CQP = 10°

(3) ѧࡵҠ∠CAP =∠CQP ☐APCQ öṺǧ ∠CAQ =∠CPQ Р∠AQP =∠ACP ∠CAQ = 10°Р∠AQP = 20°- x

(4) ˹ش R AB PR = AP ∆APR 繠∆˹Ҩ ՠ∠P ʹ ∠ARP =∠PAR ∠ARP = 20° ∠BPR = 20°- x

(5) ˹ش S AQ PS = AP ∆APS 繠∆˹Ҩ ՠ∠P ʹ ∠ASP =∠PAS ∠ASP = 20° ∠QPS = x

(6) ѧࡵҠ∆BPR∆PQS ¤ѹẺ -- (∠PBR =∠QPS, BP = PQ,∠BPR =∠PQS) BR = PS (= AP) BR = PR ∆BPR 繠∆˹Ҩ ՠ∠R ʹ ∠PBR =∠BPR x = 20°- x x = 10° Q.E.D.




Create Date : 27 áҤ 2558
Last Update : 27 áҤ 2558 0:00:01 .
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Fun Geometry Problem with Solution #153



٨ x= 20°
٨



(1)∠ACB = 30°- x,∠ADB = 60°- x Р∠BDC = 80°+ x
∵ ∠CAD =∠ACD ∆ACD 繠∆˹Ҩ ՠ∠D ʹ AD = CD

(2) ˹ش P ˹ CD CP = AB Р∠DCP = 20°+ x
Ҡ∆CDP∆ABD ¤ѹẺ -- (CD = AD,∠DCP =∠BAD, CP = AB) ∠CDP =∠ADB Р∠CPD = ∠ABD ∠CDP = 60°- x Р∠CPD = 100°
͡ҡ ѧ DP = BD ∆BDP 繠∆˹Ҩ ՠ∠D (= 140°) ʹ ∠DBP = 20°Р∠BPD = 20° ∠CBP = 30°Р∠BPC = 80°

(3) ˹ش Q Ҿз͹ͧش P ҹ BC ∆BCQ∆BCP BQ = BP,∠CBQ =∠CBP = 30°Р∠BQC = ∠BPC = 80°
∵ BP = BQ Р∠PBQ = 60° ∆BPQ 繠∆ҹ BP = PQ
͡ҡ ѧҠ∠BPQ =∠BQP = 60° ∠CPQ =∠CQP = 20°

(4) ѧࡵҠ∆BDP∆CPQ ¤ѹẺ -- (∠DBP =∠CQP, BP = PQ,∠BPD =∠CPQ) DP = CP ∆CDP ∆˹Ҩ ՠ∠P ʹ ∠DCP =∠CDP 20°+ x = 60°- x x = 20° Q.E.D.




Create Date : 24 áҤ 2558
Last Update : 24 áҤ 2558 0:00:00 .
Counter : 534 Pageviews.

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Fun Geometry Problem with Solution #152



٨ x= 10°
٨



(1)∠APB = 120°

(2) ˹ش Q AB PQ = BP ∆BPQ 繠∆˹Ҩ ՠ∠P ʹ ∠BQP =∠PBQ ∠BQP = 40° ∠APQ = 20° ∠APQ =∠PAQ ∆APQ 繠∆˹Ҩ ՠ∠Q ʹ AQ = PQ

(3) ˹ش R AP ∠PAR = 30°Р∠APR = 40° ∠QAR = 10°,∠BPR = 80°,∠QPR = 20°Р∠ARP = 110°

α= 10°
ѧࡵ ☐AQPR ՠAQ = PQ,∠QAR =α,∠QPR = 2αР∠ARP = 120°-α PR = PQ (Click ʹԸվ٨⨷ 3) PR = BP

♦∵∠ACP +∠ARP = 180° ☐ACPR öṺǧ ∠PCR =∠PAR ∠PCR = 30°

(4) ѧࡵ BP = PR Р∠BPR = 2(∠BCR) ش P circumcenter ͧ∆BCR BP = CP ∆BCP 繠∆˹Ҩ ՠ∠P ʹ ∠CBP =∠BCP x = 10° Q.E.D.




Create Date : 21 áҤ 2558
Last Update : 29 áҤ 2558 23:42:00 .
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Fun Geometry Problem with Solution #151



˹ AE = BE
٨ x= 22.5°
٨ 1



(1)∠ABC = 90°- 3x

(2) ˹ش P BC PE⊥AB
Ҡ∆AEP∆BEP ¤ѹẺ -- (AE = BE,∠AEP =∠BEP, EP = EP) ∠EAP =∠EBP ∠EAP = 90°- 3x ∠APE = 3x

(3)∵∠ACE =∠APE ☐ACPE öṺǧ ∠EAP =∠ECP 90°- 3x = x x = 22.5° Q.E.D.

٨ 2(ɮպǡѺǧ)



(1)∠ABC = 90°- 3x,∠AEC = 90°- 2x Р∠BEC = 90°+ 2x

(2) ˹ش P Ҿз͹ͧش B ҹ CE ∆CEP∆BCE ...
CP = BC
∠ECP =∠BCE ∠ECP = x ∠DCP = x
∠CEP =∠BEC ∠CEP = 90°+ 2x ∠AEP = 4x
∠CPE =∠CBE ∠CPE = 90°- 3x
EP = BE EP = AE ∆AEP 繠∆˹Ҩ ՠ∠E (= 4x) ʹ ∠APE (=∠EAP) = 90°- 2x ∠APC = x

(3) ˹ش Q 繨شѴҧ AB CP
ѧࡵҠ∆ACD ∆CDQ¤ѹẺ -- (∠ACD=∠DCQ, CD = CD,∠ADC=∠CDQ) AC= CQ

(4) ѧࡵҠ∆ACP∆BCQ ¤ѹẺ -- (AC = CQ,∠ACP =∠BCQ, CP = BC) ∠APC =∠CBQ x = 90°- 3x x = 22.5° Q.E.D.




Create Date : 18 áҤ 2558
Last Update : 18 áҤ 2558 0:00:00 .
Counter : 623 Pageviews.

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TIYHz
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