Fun Geometry Problem with Solution #133
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
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ãËé BP = L

(1) ∠APC = 120° áÅР∠BPC = 140°      ∠APB = 100°

(2) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ P ¼èÒ¹ BC      ∆BCQ  ∆BCP      BQ = BP , CQ = CP, ∠CBQ = ∠CBP = 30° áÅÐ ∠BCQ = ∠BCP = 10°
∵ BP = BQ áÅР∠PBQ = 60°      ∆BPQ à»ç¹ ∆´éÒ¹à·èÒ      PQ = BP      PQ = L

(3) ¡Ó˹´¨Ø´ R º¹ AC ·Õè·ÓãËé CR = CP      CR = CQ
¨ÐàËç¹ÇèÒ ∆CPR  ∆CPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (CP = CP, ∠PCR = ∠PCQ, CR = CQ)      PR = PQ      PR = L
∵ CP = CR      ∆CPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= 20°) à»ç¹ÁØÁÂÍ´      ∠CPR = 80° áÅР∠CRP = 80°      ∠APR = 40° áÅÐ ∠ARP = 100°
∵ ∠PAR = ∠APR      ∆APR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´      AR = PR      AR = L

(4) ¡Ó˹´¨Ø´ S à˹×Í PR ·Õè·ÓãËé PS = RS = PR (= L)      ∆PRS à»ç¹ ∆´éÒ¹à·èÒ      ∠RPS = ∠PRS = 60°
∵ AR = RS      ∆ARS à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R (= 160°) à»ç¹ÁØÁÂÍ´      ∠RAS (= ∠ASR) = 10°      ∠PAS = 30°

(5) ÊѧࡵÇèÒ ∆ABP  ∆APS ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = AP, ∠APB = ∠APS, BP = PS)      ∠BAP = ∠PAS      x = 30°   Q.E.D.

´Ù⨷Âì·Ñé§ËÁ´ Click !!



Create Date : 25 ¾ÄÉÀÒ¤Á 2558
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