Fun Geometry Problem with Solution #135
⨷Âì  ¨§¾ÔÊÙ¨¹ìÇèÒ x = 20° ¾ÔÊÙ¨¹ì  (1) ∠ACD = 90° - x, ∠ADC = 10° + x áÅÐ ∠BDC = 170° - x (2) ¡Ó˹´¨Ø´ P à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ C ¼èÒ¹ AD ⇒ ∆ADP ≅ ∆ACD ⇒ AP = AC, DP = CD, ∠DAP = ∠CAD = 80° áÅÐ ∠ADP = ∠ADC = 10° + x ∵ AC = AP ⇔ ∆ACP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A (= 160°) à»ç¹ÁØÁÂÍ´ ⇔ ∠ACP = ∠APC = 10° (3) ¡Ó˹´¨Ø´ Q ·Ò§´éÒ¹¢ÇҢͧ AC ·Õè·ÓãËé ∠ACQ = 170° - x (⇔ ∠DCQ = 80°) áÅÐ CQ = CD ¨ÐàËç¹ÇèÒ ∆ACQ ≅ ∆BCD ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AC = BD, ∠ACQ = ∠BDC, CQ = CD) ⇒ ∠AQC = ∠BCD ⇔ ∠AQC = 10° ⇔ ∠AQC = ∠APC ⇔ ☐ACQP ÊÒÁÒöṺã¹Ç§¡ÅÁä´é ⇔ ∠AQP = ∠ACP ⇔ ∠AQP = 10° ∵ CD = CQ ⇔ ∆CDQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= 80°) à»ç¹ÁØÁÂÍ´ ⇔ ∠CDQ = 50° áÅÐ ∠CQD = 50° (⇔ ∠DQP = 30°) (4) ¡Ó˹´¨Ø´ R à»ç¹¨Ø´µÑ´ÃÐËÇèÒ§ CD áÅÐ PQ ¾Ô¨ÒÃ³Ò ∆CQR ¨Ðä´éÇèÒ ∠CRQ = 80° ⇔ ∠CRQ = ∠QCR ⇔ ∆CQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ QR = CQ ⇒ QR = DP (5) ¾Ô¨ÒÃ³Ò ∆DQR ¨ÐàËç¹ÇèÒ ∠D = 50°, ∠Q = 30° áÅÐÁըش P º¹ÊèǹµèÍ¢ÂÒ¢ͧ QR «Öè§ DP = QR ⇒ ∠DPQ = 40° (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇔ ∠PDR = 60° ⇔ 20° + 2x = 60° ⇔ x = 20° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #134
⨷Âì  ¨§¾ÔÊÙ¨¹ìÇèÒ x = 20° ¾ÔÊÙ¨¹ì 1  (1) ∵ ∠BAC = ∠ABC ⇔ ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´ ⇔ AC = BC (2) ¡Ó˹´¨Ø´ Q à˹×Í AB ·Õè·ÓãËé AQ = BQ = AB ⇔ ∆ABQ à»ç¹ ∆´éÒ¹à·èÒ ⇒ ∠BAQ = 60° (⇔ ∠PAQ = 30°), ∠ABQ = 60° (⇔ ∠PBQ = 20°) áÅÐ ∠AQB = 60° ♦ ÊѧࡵÇèÒ ∆BCQ ≅ ∆ACQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (BC = AC, BQ = AQ, CQ = CQ) ⇒ ∠BQC (= ∠AQC) = (∠AQB)/2 = 30° ♦ ÊѧࡵÇèÒ ∆APQ ≅ ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = AP, ∠PAQ = ∠BAP, AQ = AB) ⇒ ∠AQP = ∠ABP ⇔ ∠AQP = 40° ⇔ ∠BQP = 20° ⇔ ∠CQP = 10° (3) ∵ ∠CBP = ∠CQP ⇔ ☐BPCQ ÊÒÁÒöṺã¹Ç§¡ÅÁä´é ⇔ ∠BCP = ∠BQP ⇔ x = 20° Q.E.D. ¾ÔÊÙ¨¹ì 2 (äÁèãªé·Äɮպ·à¡ÕèÂǡѺǧ¡ÅÁ)  (1) ∵ ∠BAC = ∠ABC ⇔ ∆ABC à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´ ⇔ AC = BC (2) ¡Ó˹´¨Ø´ Q à˹×Í AB ·Õè·ÓãËé AQ = BQ = AB ⇔ ∆ABQ à»ç¹ ∆´éÒ¹à·èÒ ⇒ ∠BAQ = 60° (⇔ ∠PAQ = 30°), ∠ABQ = 60° (⇔ ∠CBQ = 10°) áÅÐ ∠AQB = 60° ♦ ÊѧࡵÇèÒ ∆BCQ ≅ ∆ACQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (BC = AC, BQ = AQ, CQ = CQ) ⇒ ∠BQC (= ∠AQC) = (∠AQB)/2 = 30° ♦ ÊѧࡵÇèÒ ∆APQ ≅ ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = AP, ∠PAQ = ∠BAP, AQ = AB) ⇒ PQ = BP ⇔ ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠BQP = ∠PBQ ⇔ ∠BQP = 20° ⇔ ∠CQP = 10° (3) µèÍ QC ÍÍ¡ä»Âѧ¨Ø´ R â´Â·Õè ∠BRQ = 100° ⇒ ∠BCR = 40° ¾Ô¨ÒÃ³Ò ∆BQR ¨ÐàËç¹ÇèÒ ∠Q = 30°, ∠R = 100° áÅÐÁըش C º¹ QR ·Õè·ÓãËé ∠BCR = 40° ⇒ BC = QR (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 1-1) (4) ÊѧࡵÇèÒ ∆PQR ≅ ∆BCP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (PQ = BP, ∠PQR = ∠CBP, QR = BC) ⇒ ∠PRQ = ∠BCP ⇔ ∠PRQ = x ¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ PR = CP ⇔ ∆CPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠PCR = ∠CRP ⇔ ∠PCR = x (5) ∵ ∠BCP + ∠PCR = ∠BCR ⇔ x + x = 40° ⇔ x = 20° Q.E.D. ¾ÔÊÙ¨¹ì 3  (1) µèÍ BP Í͡仾º AC ·Õè¨Ø´ Q ¾Ô¨ÒÃ³Ò ∆ABQ ¨Ðä´éÇèÒ ∠AQB = 90° ⇒ BP ⊥ AC (2) ãËé α = 10° ¾Ô¨ÒÃ³Ò ∆ABC ¨ÐàËç¹ÇèÒ Áըش P à»ç¹¨Ø´ÀÒÂã¹·Õè·ÓãËé BP ⊥ AC, ∠BAP = 30°, ∠CAP = 30° - α áÅÐ ∠CBP = α ⇒ ∠BCP = 2α (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇔ x = 20° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #133
⨷Âì  ¨§¾ÔÊÙ¨¹ìÇèÒ x = 30° ¾ÔÊÙ¨¹ì  ãËé BP = L (1) ∠APC = 120° áÅÐ ∠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 áÅÐ ∠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° áÅÐ ∠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 !! Fun Geometry Problem with Solution #132
⨷Âì  ¨§¾ÔÊÙ¨¹ìÇèÒ x = 3α ¾ÔÊÙ¨¹ì  (1) ∠ACB = 60° + α (2) ¡Ó˹´¨Ø´ Q à˹×Í AB ·Õè·ÓãËé AQ = BQ = AB ⇔ ∆ABQ à»ç¹ ∆´éÒ¹à·èÒ ⇒ ... • ∠BAQ = 60° ⇔ ∠CAQ = 30° • ∠ABQ = 60° ⇒ ∠CBQ = 30° - α áÅÐ ∠PBQ = 30° • ∠AQB = 60° ♦ ÊѧࡵÇèÒ ∆ACQ ≅ ∆ABC ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AC = AC, ∠CAQ = ∠BAC, AQ = AB) ⇒ CQ = BC ⇔ ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C à»ç¹ÁØÁÂÍ´ ⇔ ∠BQC = ∠CBQ ⇔ ∠BQC = 30° - α ♦ ÊѧࡵÇèÒ ∆BPQ ≅ ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BP = BP, ∠PBQ = ∠ABP, BQ = AB) ⇒ PQ = AP ¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BQP = ∠BAP ⇔ ∠BQP = α ⇔ ∠AQP = 60° - α ¾Ô¨ÒÃ³Ò ☐ABQP ¨Ðä´éÇèÒ ∠APQ (ÁØÁãËè) = 360° - (60° + 2α) ⇔ ∠APQ (ÁØÁàÅç¡) = 60° + 2α (3) ¡Ó˹´¨Ø´ R ãµé AQ ·Õè·ÓãËé ∠QAR = 30° - α áÅÐ ∠AQR = 30° - α (⇔ ∠PQR = 30°) ¨ÐàËç¹ÇèÒ ∆AQR ≅ ∆BCQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠QAR = ∠CBQ, AQ = BQ, ∠AQR = ∠BQC) ⇒ QR = CQ ∵ ∠QAR = ∠AQR ⇔ ∆AQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´ ⇔ AR = QR ♦ ÊѧࡵÇèÒ ∆PQR ≅ ∆APR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (PQ = AP, PR = PR, QR = AR) ⇒ ∠QPR (= ∠APR) = (∠APQ)/2 = 30° + α ♦ ÊѧࡵÇèÒ ∆CPQ ≅ ∆PQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (CQ = QR, ∠CQP = ∠PQR, PQ = PQ) ⇒ ∠CPQ = ∠QPR ⇔ ∠CPQ = 30° + α (4) ¾Ô¨ÒÃ³Ò ∆ACP ¨Ðä´éÇèÒ ∠ACP = 60° - 2α ⇔ ∠BCP = x = 3α Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #131
⨷Âì  ¨§¾ÔÊÙ¨¹ìÇèÒ x = 48° ¾ÔÊÙ¨¹ì  (1) ∠ACB = 78° (2) µèÍ BC ÍÍ¡ä»Âѧ¨Ø´ Q â´Â·Õè ∠AQB = 72° ⇒ ∠CAQ = 6° ∵ ∠ABQ = ∠AQB ⇔ ∆ABQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´ ⇔ AB = AQ ÊѧࡵÇèÒ ∆APQ ≅ ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = AP, ∠PAQ = ∠BAP, AQ = AB) ⇒ PQ = BP ⇔ ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠BQP = ∠PBQ ⇔ ∠BQP = 36° ⇔ ∠BPQ = 108° (3) ¡Ó˹´¨Ø´ R à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ C ¼èÒ¹ AB ⇒ ∆ABR ≅ ∆ABC ⇒ AR = AC, BR = BC, ∠BAR = ∠BAC = 30° áÅÐ ∠ARB = ∠ACB = 78° ∵ AC = AR áÅÐ ∠CAR = 60° ⇒ ∆ACR à»ç¹ ∆´éÒ¹à·èÒ ⇒ AC = CR áÅÐ ∠ACR = ∠ARC = 60° (⇔ ∠BCR = ∠BRC = 18°) (4) ¡Ó˹´¨Ø´ S à˹×Í AC ·Õè·ÓãËé ∠CAS = ∠ACS = 18° (⇒ ∠ASC = 144°) ¨ÐàËç¹ÇèÒ ∆ACS ≅ ∆BCR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠CAS = ∠BCR, AC = CR, ∠ACS = ∠BRC) ⇒ AS = BC áÅÐ CS = BR ⇒ AS = BC = CS (5) ∵ AS = CS áÅÐ ∠ASC = 2(∠AQC) ⇒ ¨Ø´ S à»ç¹ circumcenter ¢Í§ ∆ACQ ⇒ ∠CSQ = 2(∠CAQ) ⇔ ∠CSQ = 12° ∵ BC = CS ⇔ ∆BCS à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= 96°) à»ç¹ÁØÁÂÍ´ ⇔ ∠BSC (= ∠CBS) = 42° ∵ BP = PQ áÅÐ ∠BPQ = 2(∠BSQ) ⇒ ¨Ø´ P à»ç¹ circumcenter ¢Í§ ∆BQS ⇒ BP = PS (6) ÊѧࡵÇèÒ ∆BCP ≅ ∆CPS ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (BC = CS, BP = PS, CP = CP) ⇒ ∠BCP (= ∠PCS) = (∠BCS)/2 ⇔ x = 48° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! |
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