標題:
幾何ｘ２(3)
發問:
1. 已知：角BAD = 角CBE = 角ACF，角FDE = 60度，角DEF = 45度。 求三角形ABC各內角的度數。 2.已知：三角形ABC中，AB = AC，D在BC上，角BAD = 30度，在AC上截取AE = AD。 求角EDC的度數。 圖在： http://www.xanga.com/highlowb (pic1：第1題；pic2：第2題)
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最佳解答:
(1) Let ∠ BAD be y. ∠ BAD = ∠ CBE = ∠ ACF = y ( given ) ∠ DEF + ∠ FDE + ∠ DFE = 180 * ( ∠ sum of △ ) 45* + 60* + ∠ DFE = 180* ∠ DFE = 75* Let ∠ ABD = x , ∠ BEC = z, ∠ FAC = a x + y = 60* ( ext ∠ of △ ) z + y = 45* ( ext ∠ of △ ) a + y = 75* ( ext ∠ of △ ) So ∠ ABC = x + y = 60* , ∠ BCA = z + y = 45*, ∠ CAB = a + y = 75* (2) Let ∠ ADE be x and ∠ EDC be y. ∠ ADE = ∠ AED = x ( base ∠s , isos. △ ) 30* + ∠ ABD = x + y ( ext. ∠s of △ ) ∠ ABD = x + y - 30* ∠ ABD = ∠ ACB = x + y - 30* ( base ∠s , isos. △ ) In △ DEC, y + ( x + y - 30* ) = x ( ext. ∠s of △ ) 2y = 30* y = 15* Therefore ∠ EDC is 15* 2007-06-25 19:04:25 補充： It should be ( ext. ∠ of △ ) but not ( ext. ∠s of △ ), sorry.
其他解答:
1. ∠ABD=∠FDE-∠BAD (ext. ∠of△) ∠ABD=60°-∠BAD ∠ABC=∠ABD+∠CBE (given) ∠ABC=60°-∠BAD+∠CBE ∵∠BAD=∠CBE (given) ∴∠ABC=60°-∠BAD+∠BAD ∴∠ABC=60° ∠BCE=∠DEF-∠CBD (ext. ∠of△) ∠BCE=45°-∠CBD ∠ACB=∠BCE+∠ACF (given) ∠ACB=45°-∠CBD+∠ACF ∵∠CBD=∠ACF (given) ∴∠ACB=45°-∠CBD+∠CBD ∴∠ACB=45° ∠ABC+∠ACB+∠BAC=180°(∠ sum of △) ∠BAC=180°-60°-45° ∠BAC=75° 2. Consider △ADC ∵AB=AC (given) ∴∠ABC=∠ACB (base ∠s,isos.△) Consider △ABD ∠ABC+∠BAD+∠ADB=180° (∠ sum of △) ∠ABC+30°+∠ADB=180° ∠ADB=180°-30°-∠ABC ∠ADB=150°-∠ABC ∵∠ABC=∠ACB (proved) ∴∠ADB=150-∠ACB Consider △ADE ∵AD=AE (given) ∴∠ADE=∠AED (base ∠s,isos.△) Consider △DEC ∠AED=∠EDC+∠ACB (ext. ∠ of △) ∵∠ADE=∠AED(proved) ∴∠ADE=∠EDC+∠ACB Consider straight line BDC ∠ADB+∠ADE+∠EDC=180° (adj. ∠s on st. line) 150°-∠ACB+∠EDC+∠ACB+∠EDC=180° 150°-2x∠EDC=180° 2x∠EDC=30° ∠EDC=15°
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