Fun Geometry Problem with Solution #149



٨ x= 10°
٨ 1



AD = L

(1)∠ACB = 100°Р∠ADB = 40°

(2) ˹ش P Ҿз͹ͧش D ҹ AB ∆ABP∆ABD BP = BD,∠ABP =∠ABD = 30°Р∠APB =∠ADB = 40°
∵ BD = BP Р∠DBP = 60° ∆BDP 繠∆ҹ BD = DP
͡ҡ ѧҠ∠BDP =∠BPD = 60° ∠ADP =∠APD = 20°

(3) ˹ش Q AC DQ = L DQ = AD ∆ADQ 繠∆˹Ҩ ՠ∠D ʹ ∠AQD =∠DAQ ∠AQD = 80° ∠ADQ = 20° ∠BDQ = 20°

(4) ѧࡵҠ∆BDQ∆ADP ¤ѹẺ -- (BD = DP,∠BDQ =∠ADP, DQ = AD) ∠DBQ =∠APD ∠DBQ = 20° ∠DBQ =∠BDQ ∆BDQ 繠∆˹Ҩ ՠ∠Q ʹ BQ = DQ BQ = L
ԨóҠ∆BCQ Ҡ∠BQC = 40°

(5) ԨóҠ∆ABC Ҡ∠A = 30°,∠C = 100°ըش Q AC ∠BQC = 40° AC = BQ (Click ʹԸվ٨⨷ 1-1) AC = L AC = AD ∆ACD 繠∆˹Ҩ ՠ∠A (= 80°) ʹ ∠ADC (=∠ACD) = 50° ∠BDC = x = 10° Q.E.D.

٨ 2



(1)∠ACB = 100°Р∠ADB = 40°

(2) ˹ش P 繨شѴҧ AC BD
∵ ∠BAP =∠ABP ∆ABP 繠∆˹Ҩ ՠ∠P ʹ AP = BP

(3) ˹ش Q AD PQ = AP ∆APQ 繠∆˹Ҩ ՠ∠P ʹ ∠AQP =∠PAQ ∠AQP = 80° ∠APQ = 20°Р∠DPQ = 40°
∵ ∠PDQ =∠DPQ ∆DPQ 繠∆˹Ҩ ՠ∠Q ʹ DQ = PQ DQ = AP

(4) ˹ش R PQ AR = AQ ∆AQR 繠∆˹Ҩ ՠ∠A ʹ ∠ARQ =∠AQR ∠ARQ = 80° ∠ARP = 100°
ѧࡵҠ∆BCP∆APR ¤ѹẺ -- (∠BCP =∠ARP,∠CBP =∠APR, BP = AP) CP = AR CP = AQ

(5)∵AP = DQ CP = AQ AP + CP = DQ + AQ AC = AD ∆ACD 繠∆˹Ҩ ՠ∠A (= 80°) ʹ ∠ADC (=∠ACD) = 50° ∠BDC = x = 10° Q.E.D.




Create Date : 12 áҤ 2558
Last Update : 12 áҤ 2558 0:00:00 .
Counter : 504 Pageviews.

0 comments
:
Comment :
 * code html 觢ͤ੾Ҫԡ
 

TIYHz
Location :
ا෾  Thailand

[ Profile ]
ԻҢͧ Blog [?]
 ҡͤѧ
 Rss Feed
 Smember
 Դ͡ : 20 [?]



شʧӺ͡Եʵ... ҡѺ ҡẺҧ
All Blog