⨷Âì
¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì 1
(1) ¡Ó˹´¨Ø´ P ·Õè·ÓãËé ☐ABPD à»ç¹ ☐¨ÑµØÃÑÊ ⇒ AB = AD = BP = DP ⇔ BC = BP = DP
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠DBP = 45° ⇔ ∠CBP = 60°
(2) ∵ BC = BP áÅÐ ∠CBP = 60° ⇒ ∆BCP à»ç¹ ∆´éÒ¹à·èÒ ⇒ BP = CP áÅÐ ∠BPC = 60°
ÊѧࡵÇèÒ BP = CP = DP ⇔ ¨Ø´ P à»ç¹¨Ø´ÈÙ¹Âì¡ÅÒ§¢Í§Ç§¡ÅÁ·ÕèÁÕ ∆BCD Ṻ㹠⇒ ∠BDC = (∠BPC)/2 ⇔ x = 30° Q.E.D.
¾ÔÊÙ¨¹ì 2
ãËé AB (= AD = BC) = a
(1) ¡Ó˹´¨Ø´ P áÅШش Q ·Õè·ÓãËé ☐ADQP à»ç¹ ☐ÁØÁ©Ò¡ â´ÂÁըش C ÍÂÙ躹 PQ ⇒ ...
• AP = DQ
• PQ = AD ⇔ PQ = a
• ∠ADQ = ∠APQ = ∠DQP = 90°
(2) ∵ AB = AD ⇔ ∆ABD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A (= 90°) à»ç¹ÁØÁÂÍ´ ⇔ ∠ABD = 45° (⇔ ∠CBP = 30°) áÅÐ ∠ADB = 45° (⇔ ∠BDQ = 45°)
∵ AB = BC ⇔ ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 150°) à»ç¹ÁØÁÂÍ´ ⇔ ∠BAC (= ∠ACB) = 15°
(3) ¨ÐàËç¹ÇèÒ ∆BCP à»ç¹ ∆ÁØÁ©Ò¡ ·ÕèÁÕ ∠P à»ç¹ÁØÁ©Ò¡ áÅÐ ∠B = 30° ⇒ CP = BC/2 ⇔ CP = a/2 ⇔ CQ = a/2
ÊѧࡵÇèÒ ∆CDQ ≅ ∆ACP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (DQ = AP, ∠CQD = ∠APC, CQ = CP) ⇒ ∠CDQ = ∠CAP ⇔ ∠CDQ = 15° ⇔ ∠BDC = x = 30° Q.E.D.
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