Proof
22
Proof D and E are the middle points of AB and AC, DE intersects the circle at G and F, CF // AB. Prove: 1.CD = BC; 2.ΔBCD ~ ΔGBD
description
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. - PowerPoint PPT Presentation
Transcript of Proof
Proof
D and E are the middle points of AB and AC, DE intersects the circle at G and F, CF // AB.Prove: 1. CD = BC;2. ΔBCD ~ ΔGBD
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
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // AB
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogram
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = AD
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = ADConnect AF, Since AD // CF
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = ADConnect AF, Since AD // CF ADCF is a parallelogram
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = ADConnect AF, Since AD // CF ADCF is a parallelogramSo CD = AF
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = ADConnect AF, Since AD // CF ADCF is a parallelogramSo CD = AFBecause CF // AB
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = ADConnect AF, Since AD // CF ADCF is a parallelogramSo CD = AFBecause CF // ABSo AF = BC
Prove: 1. CD = BC
Because AD = DB, AE = EC Then DE // BCWe knew CF // ABSo BCFD is a parallelogramThen CF = BD = ADConnect AF, Since AD // CF ADCF is a parallelogramSo CD = AFBecause CF // ABSo AF = BCThus CD = BC, since AF = DC
Prove: 2. ΔBCD ~ ΔGBD
Prove: 2. ΔBCD ~ ΔGBD
Because FG // BC
Prove: 2. ΔBCD ~ ΔGBD
Because FG // BCThen GB = CF, and GB = BD
Prove: 2. ΔBCD ~ ΔGBD
Because FG // BCThen GB = CF, and GB = BDSo DGB = EFC = DBC∠ ∠ ∠
Prove: 2. ΔBCD ~ ΔGBD
Because FG // BCThen GB = CF, and GB = BDSo DGB = EFC = DBC∠ ∠ ∠Because GB = BD, and We proved
CD = BC in the first part
Prove: 2. ΔBCD ~ ΔGBD
Because FG // BCThen GB = CF, and GB = BDSo DGB = EFC = DBC∠ ∠ ∠Because GB = BD, and We proved
CD = BC in the first partSo DGB = GDB = DBC∠ ∠ ∠ = BDC∠
Prove: 2. ΔBCD ~ ΔGBD
Because FG // BCThen GB = CF, and GB = BDSo DGB = EFC = DBC∠ ∠ ∠Because GB = BD, and We proved
CD = BC in the first partSo DGB = GDB = DBC∠ ∠ ∠ = BDC∠So ΔBCD ~ ΔGBD
Citation"2012 College Entrance Examination: Math Exam." College Entrance
Exams. Zhong Guo Jiao Yu Xin Wen Wang, 3 June 2012. Web. 22 Apr. 2013. <http://www.jyb.cn/gk/gkrdzt/2012gkstda/>
Thank You!
![arXiv:0807.4814v1 [math-ph] 30 Jul 2008 · 4.7. Proof of Theorem 2.3 and proof of Proposition 2.4 34 4.8. Proof of Proposition 2.6 34 5. A Riemann surface 34 5.1. A four-sheeted Riemann](https://static.fdocument.org/doc/165x107/5f72beda42271e6b1d0c087c/arxiv08074814v1-math-ph-30-jul-2008-47-proof-of-theorem-23-and-proof-of-proposition.jpg)
