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Urgent!! geometry problem
Refer to the picture: http://img96.imageshack.us/img96/8840/questionlf9.png A right-angled traingle ABC,where angle ACB=90 dregee and CD perpendicular AB.ADB is a straight line.If traingle ACE and traingle BFC are equilateral traingle,prove that traingle ADE similar traingle CDF.
最佳解答:
Let ∠BCD be θ, then ∠ACD = 90度 - θ and hence ∠CAD = θ Also, ∠BCF = ∠DAE = 60度 and hence: ∠DCF = ∠DAE = 60度 + θ Now, looking into △BCD and △CAD: ∠BDC = ∠CDA = 90度 ∠BCD = ∠CAD = θ Hence △BCD ~ △CAD and then their corresponding side ratios are equal, i.e. CA/BC = AD/CD = DC/DB --- (1) Moreover, since △ACE and △FBC are equilateril triangles, CA = AE and BC = CF and therefore: CA/BC = AE/CF --- (2) Combining (1) and (2), we have: AD/CD = CA/BC = AE/CF AD/CD = AE/CF With 2 corresponding side ratios equal and together with the same included angle (∠DCF = ∠DAE), the conclusion is: △ADE ~ △CDF.
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Urgent!! geometry problem
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發問:Refer to the picture: http://img96.imageshack.us/img96/8840/questionlf9.png A right-angled traingle ABC,where angle ACB=90 dregee and CD perpendicular AB.ADB is a straight line.If traingle ACE and traingle BFC are equilateral traingle,prove that traingle ADE similar traingle CDF.
最佳解答:
Let ∠BCD be θ, then ∠ACD = 90度 - θ and hence ∠CAD = θ Also, ∠BCF = ∠DAE = 60度 and hence: ∠DCF = ∠DAE = 60度 + θ Now, looking into △BCD and △CAD: ∠BDC = ∠CDA = 90度 ∠BCD = ∠CAD = θ Hence △BCD ~ △CAD and then their corresponding side ratios are equal, i.e. CA/BC = AD/CD = DC/DB --- (1) Moreover, since △ACE and △FBC are equilateril triangles, CA = AE and BC = CF and therefore: CA/BC = AE/CF --- (2) Combining (1) and (2), we have: AD/CD = CA/BC = AE/CF AD/CD = AE/CF With 2 corresponding side ratios equal and together with the same included angle (∠DCF = ∠DAE), the conclusion is: △ADE ~ △CDF.
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