Fun Geometry Problem with Solution #145
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¨§¾ÔÊÙ¨¹ìÇèÒ α + β + γ = 90°
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(1) ∵ AB = BC      ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 90°) à»ç¹ÁØÁÂÍ´      ∠ACB (= ∠BAC) = 45°   ⇔   α = 45°

(2) ¨Ò¡ÃÙ» ¨ÐàËç¹ÇèÒ ∆APQ  ∆EQR  ∆ABD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = QR = AB, ∠APQ = ∠ERQ = ∠ABD, PQ = ER = BD)      ∠AQP = ∠QER = ∠ADB = β

(3) ¾Ô¨ÒóҠ∆EQR ¨Ðä´éÇèÒ ∠EQR = 90° - β
∵ ∠AQP + ∠AQE + ∠EQR = 180°      β + ∠AQE + (90° - β) = 180°      ∠AQE = 90°

(4) ∵ ∆APQ  ∆EQR      AQ = EQ      ∆AEQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q (= 90°) à»ç¹ÁØÁÂÍ´      ∠AEQ (= ∠EAQ) = 45°      ∠AEQ = α
∵ ∠AEQ + ∠QER + ∠AEB = 90°   ⇔   α + β + γ = 90°   Q.E.D.

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Fun Geometry Problem with Solution #144
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¡Ó˹´ãËé ∠BCD à»ç¹ÁØÁ»éÒ¹
¨§¾ÔÊÙ¨¹ìÇèÒ x = 15°
¾ÔÊÙ¨¹ì



(1) ¾Ô¨ÒóҠ∆ABD ¨Ðä´éÇèÒ ∠ADB = 90° - x
¾Ô¨ÒóҠ∆ACD ¨Ðä´éÇèÒ ∠CAD = 60° - x

(2) µèÍ BC ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè ∠BPD = 90°      ∠DCP = 3x
¨ÐàËç¹ÇèÒ ∆BDP  ∆ABD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-Á-´ (∠BPD = ∠BAD, ∠DBP = ∠ABD, BD = BD)      BP = AB      ∆ABP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 2x) à»ç¹ÁØÁÂÍ´      ∠BAP = ∠APB = 90° - x      ∠DAP = ∠APD = x

(3) ¡Ó˹´¨Ø´ Q à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ A ¼èÒ¹ CD      ∆CDQ  ∆ACD      CQ = AC, ∠DCQ = ∠ACD = 30° áÅР∠CQD = ∠CAD = 60° - x
∵ AC = CQ áÅР∠ACQ = 60°      ∆ACQ à»ç¹ ∆´éÒ¹à·èÒ      AC = AQ
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠CAQ = ∠AQC = 60°      ∠DAQ = ∠AQD = x

(4) ÊѧࡵÇèÒ ∆ADP  ∆ADQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-Á-´ (∠APD = ∠AQD, ∠DAP = ∠DAQ, AD = AD)      AP = AQ      AP = AC      ∆ACP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´      ∠ACP = ∠APC      ∠ACP = 90° - x      ∠DCP = 60° - x      3x = 60° - x      x = 15°   Q.E.D.

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Create Date : 27 ÁԶعÒ¹ 2558
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Fun Geometry Problem with Solution #143
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 40° â´ÂäÁèãªé·Äɮպ·à¡ÕèÂǡѺǧ¡ÅÁ
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(1) µèÍ AP Í͡仾º BC ·Õè¨Ø´ Q      ∠BPQ = 40°, ∠AQB = 100° áÅР∠AQC = 80°
∵ ∠BPQ = ∠PBQ      ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´      BQ = PQ

(2) ¡Ó˹´¨Ø´ R à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ Q ¼èÒ¹ AC      ∆ACR  ∆ACQ      AR = AQ, ∠CAR = ∠CAQ = 30° áÅÐ ∠ARC = ∠AQC = 80°
∵ AQ = AR áÅР∠QAR = 60°      ∆AQR à»ç¹ ∆´éÒ¹à·èÒ      AQ = QR
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠AQR = ∠ARQ = 60°      ∠CQR = ∠CRQ = 20°

(3) ¡Ó˹´¨Ø´ S º¹ AB ·Õè·ÓãËé ∠AQS = 20°      ∠BQS = 80° áÅР∠BSQ = 40°
¨ÐàËç¹ÇèÒ ∆AQS  ∆CQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠QAS = ∠CRQ, AQ = QR, ∠AQS = ∠CQR)      QS = CQ

(4) ÊѧࡵÇèÒ ∆CPQ  ∆BQS ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (CQ = QS, ∠CQP = ∠BQS, PQ = BQ)      ∠PCQ = ∠BSQ      x = 40°   Q.E.D.

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Create Date : 24 ÁԶعÒ¹ 2558
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Fun Geometry Problem with Solution #142
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
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ãËé AP = L

(1) µèÍ AB ÍÍ¡ä»Âѧ¨Ø´ Q â´Â·Õè PQ = L      ∠CBQ = 140°
∵ AP = PQ      ∆APQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠AQP = ∠PAQ      ∠AQP = 10°

(2) µèÍ AC ÍÍ¡ä»Âѧ¨Ø´ R â´Â·Õè PR = L      ∠BCR = 70°
∵ AP = PR      ∆APR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      ∠ARP = ∠PAR      ∠ARP = 20°
¾Ô¨ÒóҠ∆CPR ¨Ðä´éÇèÒ ∠CPR = 80°      ∠CPR = ∠PCR      ∆CPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´      CR = PR      CR = L

(3) ¾Ô¨ÒóҠ☐AQPR ¨Ðä´éÇèÒ ∠QPR (ÁØÁãË­è) = 360° - 60°      ∠QPR (ÁØÁàÅç¡) = 60°
∵ PQ = PR áÅР∠QPR = 60°      ∆PQR à»ç¹ ∆´éÒ¹à·èÒ      QR = L áÅР∠PQR = ∠PRQ = 60°
∵ CR = QR      ∆CQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R (= 80°) à»ç¹ÁØÁÂÍ´      ∠QCR = ∠CQR = 50°      ∠BCQ = ∠BQC = 20°      ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 140°) à»ç¹ÁØÁÂÍ´      BC = BQ

(4) ÊѧࡵÇèÒ ∆BQR  ∆BCR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (BQ = BC, BR = BR, QR = CR)      ∠BRQ (= ∠BRC) = (∠CRQ)/2 = 40°      ∠BRP = 20°
¾Ô¨ÒóҠ∆BQR ¨Ðä´éÇèÒ ∠QBR = 70° ( ∠ABR = 110°)      ∠QBR = ∠BQR      ∆BQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´      BR = QR   ⇔   BR = L
∵ BR = PR      ∆BPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R (= 20°) à»ç¹ÁØÁÂÍ´      ∠PBR (= ∠BPR) = 80°      ∠ABP = x = 30°   Q.E.D.

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Create Date : 21 ÁԶعÒ¹ 2558
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Fun Geometry Problem with Solution #141
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¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
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(1) ∠ACB = 120° - 2α áÅР∠ADB = 60° - α

(2) µèÍ DA ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè AP = AC      ∠BAP = 60°
¨ÐàËç¹ÇèÒ ∆ABP  ∆ABC ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AB = AB, ∠BAP = ∠BAC, AP = AC)      BP = BC
¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠APB = ∠ACB      ∠APB = 120° - 2α

(3) µèÍ DP ÍÍ¡ä»Âѧ¨Ø´ Q ·Õè·ÓãËé ∠BQD = 60° - α      ∠BQD = ∠BDQ      ∆BDQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´      BD = BQ
¾Ô¨ÒóҠ∆BPQ ¨Ðä´éÇèÒ ∠PBQ = 60° - α áÅР∠BPQ = 60° + 2α
∵ ∠PBQ = ∠BQP      ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´      PQ = BP      PQ = BC

(4) ¡Ó˹´¨Ø´ R ·Ò§´éÒ¹¢ÇҢͧ PQ ·Õè·ÓãËé PR = QR = PQ (= BC)      ∆PQR à»ç¹ ∆´éÒ¹à·èÒ      ∠QPR = 60° áÅР∠PQR = 60°      ∠BPR = 2α áÅР∠BQR = α
∵ BP = PR      ∆BPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P (= 2α) à»ç¹ÁØÁÂÍ´      ∠PBR (= ∠BRP) = 90° - α      ∠QBR = 30°

(5) ÊѧࡵÇèÒ ∆BCD  ∆BQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BC = QR, ∠CBD = ∠BQR, BD = BQ)      ∠BDC = ∠QBR      x = 30°   Q.E.D.

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Create Date : 18 ÁԶعÒ¹ 2558
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