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Proof

Proof. D and E are the middle points of AB and AC, DE intersects the circle at G and F, CF // AB. Prove: CD = BC; Δ BCD ~ Δ GBD. Prove: 1. CD = BC. Prove: 1. CD = BC Because AD = DB, AE = EC. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC.

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Proof

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  1. Proof D and E are the middle points of AB and AC, DE intersects the circle at G and F, CF // AB. Prove: CD = BC; ΔBCD ~ ΔGBD

  2. Prove: 1. CD = BC

  3. Prove: 1. CD = BC Because AD = DB, AE = EC

  4. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC

  5. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB

  6. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram

  7. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD

  8. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD Connect AF, Since AD // CF

  9. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD Connect AF, Since AD // CF ADCF is a parallelogram

  10. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD Connect AF, Since AD // CF ADCF is a parallelogram So CD = AF

  11. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD Connect AF, Since AD // CF ADCF is a parallelogram So CD = AF Because CF // AB

  12. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD Connect AF, Since AD // CF ADCF is a parallelogram So CD = AF Because CF // AB So AF = BC

  13. Prove: 1. CD = BC Because AD = DB, AE = EC Then DE // BC We knew CF // AB So BCFD is a parallelogram Then CF = BD = AD Connect AF, Since AD // CF ADCF is a parallelogram So CD = AF Because CF // AB So AF = BC Thus CD = BC, since AF = DC

  14. Prove: 2. ΔBCD ~ ΔGBD

  15. Prove: 2. ΔBCD ~ ΔGBD Because FG // BC

  16. Prove: 2. ΔBCD ~ ΔGBD Because FG // BC Then GB = CF, and GB = BD

  17. Prove: 2. ΔBCD ~ ΔGBD Because FG // BC Then GB = CF, and GB = BD So ∠DGB = ∠EFC = ∠DBC

  18. Prove: 2. ΔBCD ~ ΔGBD Because FG // BC Then GB = CF, and GB = BD So ∠DGB = ∠EFC = ∠DBC Because GB = BD, and We proved CD = BC in the first part

  19. Prove: 2. ΔBCD ~ ΔGBD Because FG // BC Then GB = CF, and GB = BD So ∠DGB = ∠EFC = ∠DBC Because GB = BD, and We proved CD = BC in the first part So ∠DGB = ∠GDB = ∠DBC = ∠BDC

  20. Prove: 2. ΔBCD ~ ΔGBD Because FG // BC Then GB = CF, and GB = BD So ∠DGB = ∠EFC = ∠DBC Because GB = BD, and We proved CD = BC in the first part So ∠DGB = ∠GDB = ∠DBC = ∠BDC So ΔBCD ~ ΔGBD

  21. Citation "2012 College Entrance Examination: Math Exam." College Entrance Exams. ZhongGuo Jiao Yu Xin Wen Wang, 3 June 2012. Web. 22 Apr. 2013. <http://www.jyb.cn/gk/gkrdzt/2012gkstda/>

  22. Thank You!

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