Fun Geometry Problem with Solution #130



˹ BE = AC + CE
٨ x = 45°
٨



AC = a CE = b BE = a + b

(1)∵∆ABC 繠∆ҡ Шش D 繨ش觡ҧҹçҡ (AB) AD = BD = CD
∵ AD = CD ∆ACD 繠∆˹Ҩ ՠ∠D ʹ
∵ BD = CD ∆BCD 繠∆˹Ҩ ՠ∠D ʹ

(2) ˹ش P AC DP⊥AC DP ǹ٧ͧ∆ACD CP (= AP) = AC/2 = a/2

(3) ˹ش Q BC DQ⊥BC DQ ǹ٧ͧ∆BCD CQ (= BQ) = BC/2 = a/2 + b EQ = a/2

(4) ԨóҠ☐CPDQ Ҡ∠C =∠P =∠Q = 90° ☐CPDQ 繠☐ҡ DQ = CP DQ = a/2 DQ = EQ ∆DEQ 繠∆˹Ҩ ՠ∠Q (= 90°) ʹ ∠DEQ (=∠EDQ) = x = 45° Q.E.D.




Create Date : 16 Ҥ 2558
Last Update : 16 Ҥ 2558 0:00:00 .
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Fun Geometry Problem with Solution #129



٨ x = 10°
٨(¤س Javi Master Jans)



(1)∠ACB = 60°Р∠ADB = 40°
∵ ∠BAD =∠ABD ∆ABD 繠∆˹Ҩ ՠ∠D ʹ AD = BD

(2) ˹ش P AB CP = BC ∆BCP 繠∆˹Ҩ ՠ∠C ʹ ∠BPC =∠CBP ∠BPC = 80° ∠ACP = 40°Р∠APC = 100°
∵ ∠CAP =∠ACP ∆ACP 繠∆˹Ҩ ՠ∠P ʹ AP = CP AP = BC

(3) ˹ش Q ˹ AP AQ = PQ = AP (= BC) ∆APQ 繠∆ҹ ∠PAQ = 60°Р∠APQ = 60° ∠DAQ = 10°Р∠CPQ = 40°
∵ CP = PQ ∆CPQ 繠∆˹Ҩ ՠ∠P (= 40°) ʹ ∠PCQ =∠CQP = 70°

(4) ѧࡵҠ∆ADQ∆BCD ¤ѹẺ -- (AD = BD,∠DAQ =∠CBD, AQ = BC) ∠ADQ =∠BDC ∠ADQ = x ∠BDQ = 40°- x
͡ҡ ѧ DQ = CD ∆CDQ 繠∆˹Ҩ ՠ∠D (= 40°) ʹ ∠DCQ =∠CQD = 70°

(5) ѧࡵҠ∆CDQ∆CPQ ¤ѹẺ -- (∠DCQ =∠PCQ, CQ = CQ,∠CQD =∠CQP) CD = CP CD = BC ∆BCD 繠∆˹Ҩ ՠ∠C ʹ ∠BDC =∠CBD x = 10° Q.E.D.




Create Date : 13 Ҥ 2558
Last Update : 13 Ҥ 2558 0:00:00 .
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Fun Geometry Problem with Solution #128



˹ AC = BD + CD
٨ x = 40°
٨



(1)∠CAD = x = 60°-α

(2) CD ͡ѧش P · DP = BD ∠BDP = 120°
∵ BD = DP ∆BDP 繠∆˹Ҩ ՠ∠D (= 120°) ʹ ∠BPD (=∠DBP) = 30°
∵ CP = CD + DP CP = CD + BD CP = AC ∆ACP 繠∆˹Ҩ ՠ∠C (=α) ʹ ∠CAP (=∠APC) = 90°-α/2 ∠BAP = 30°+α/2

(3) ˹ش Q Ҿз͹ͧش C ҹ BP ∆BPQ∆BCP BQ = BC, PQ = CP,∠BPQ =∠BPC = 30° ∠BQP = ∠BCP =α
∵ CP = PQ Р∠CPQ = 60° ∆CPQ 繠∆ҹ CP = CQ Р∠PCQ = 60°

(4) ѧࡵ AC = CP = CQ ش C 繨شٹҧͧǧՠ∆APQ Ṻ ∠PAQ = (∠PCQ)/2 ∠PAQ = 30° ∠BAQ =α/2
͡ҡ ѧҠ∠AQP = (∠ACP)/2 ∠AQP =α/2 ∠AQB =α/2 ∠AQB =∠BAQ ∆ABQ 繠∆˹Ҩ ՠ∠B ʹ AB = BQ AB = BC ∆ABC 繠∆˹Ҩ ՠ∠B ʹ ∠BAC =∠ACB 60°-α= 2α α= 20° x = 40° Q.E.D.




Create Date : 10 Ҥ 2558
Last Update : 22 Ҥ 2558 1:38:00 .
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Fun Geometry Problem with Solution #127



٨ x = 18°
٨



(1)∠ACD = 30°- x Р∠BCD = x

(2) ˹ش O 繨شٹҧͧǧՠ∆BCD Ṻ ...
BO = CO = DO
∠BOC = 2(∠BDC) ∠BOC = 60°
∠BOD = 2(∠BCD) ∠BOD = 2x
∵ BO = CO Р∠BOC = 60° ∆BCO 繠∆ҹ CO = BC DO = AD
∵ CO = DO ∆CDO 繠∆˹Ҩ ՠ∠O (= 60°+ 2x) ʹ ∠DCO = 60°- x (∠ACO = 30°) Р∠CDO = 60°- x

(3) ѧࡵҠ∆ACO∆ABC ¤ѹẺ -- (AC = AC,∠ACO =∠ACB, CO = BC) ∠CAO =∠BAC ∠CAO = x
∵ AD = DO ∆ADO 繠∆˹Ҩ ՠ∠D ʹ ∠AOD =∠DAO ∠AOD = 2x
ԨóҠ∆ADO Ҡ∠DAO +∠AOD =∠BDO 2x + 2x = 90°- x x = 18° Q.E.D.




Create Date : 07 Ҥ 2558
Last Update : 7 Ҥ 2558 0:00:00 .
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Fun Geometry Problem with Solution #126



٨ x = 70°
٨



(1) CB ͡ѧش P ·∠BAP = 30°
ԨóҠ∆ABC Ҡ∠ABP = 70°
ԨóҠ☐ADCP Ҡ∠DAP +∠DCP = 180° ☐ADCP öṺǧ ∠ADP =∠ACP ∠ADP = 60°

(2)∵∠DAP =∠ADP = 60° ∆ADP 繠∆ҹ AD = AP
ѧࡵҠ∆ABD∆ABP ¤ѹẺ -- (AB = AB,∠BAD =∠BAP, AD = AP) ∠ABD =∠ABP x = 70° Q.E.D.




Create Date : 04 Ҥ 2558
Last Update : 4 Ҥ 2558 0:02:00 .
Counter : 567 Pageviews.

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TIYHz
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