Question

cho điểm E thuộc cạnh bên BC cua hinh thang ABCD vẽ đường đi qua C và song song voi AE cắt AD ở K . Chứng minh BK//DE

Answer 1

Kéo dài CK,DE cắt AB lần lượt ở M,N.
*Áp dụng hệ quả đ/l Thales, ta có:
 BMC có AE//CM (gt)  ABAM=BEECABAM=BEEC (1)
 BNE có BN//DC  BNDC=BEECBNDC=BEEC (2)
 AMK có AM//DC  AMDC=AKDKAMDC=AKDK (3)
*(1),(2) suy ra ABAM=BNDCABAM=BNDC
 ABBN=AMDCABBN=AMDC, lại có (3)
ABBN=AKDKABBN=AKDK 
 Q.E.D 

Answer 2

ABAM=BEEC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">ABAM=BEECBNDC=BEEC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">BNDC=BEECAMDC=AKDK" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">AMDC=AKDKABAM=BNDC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">ABAM=BNDC
ABBN=AMDC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">ABBN=AMDCABBN=AKDK" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">ABBN=AKDK 

tích nha 
 Q.E.D 

Answer 3

ABAM=BEEC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">ABAM=BEECBNDC=BEEC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">BNDC=BEECAMDC=AKDK" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">AMDC=AKDKABAM=BNDC" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvea,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">ABAM=BNDC

Answer 4

Kéo dài CK,DE cắt AB lần lượt ở M,N.
*Áp dụng hệ quả đ/l Thales, ta có:
\(\in\) BMC có AE//CM (gt)\(\Rightarrow\frac{AB}{AM}=\frac{BE}{EC}\)
\(\in\) BNE có BN//DC  \(\Rightarrow\frac{BN}{DC}=\frac{BE}{EC}\)(2)
 AMK có AM//DC \(\Rightarrow\frac{AM}{DC}=\frac{AK}{DK}\)
*(1),(2) suy ra \(\frac{AB}{AM}=\frac{BN}{DC}\)
Chứng minh các đường thẳng song song - Hình học - Diễn đàn Toán học

Answer 5

Kéo dài CK,DE cắt AB lần lượt ở M,N.
*Áp dụng hệ quả đ/l Thales, ta có:
 BMC có AE//CM (gt)  ABAM=BEECABAM=BEEC (1)
 BNE có BN//DC  BNDC=BEECBNDC=BEEC (2)
 AMK có AM//DC  AMDC=AKDKAMDC=AKDK (3)
*(1),(2) suy ra ABAM=BNDCABAM=BNDC
 ABBN=AMDCABBN=AMDC, lại có (3)
ABBN=AKDKABBN=AKDK 
 Q.E.D 

Answer 6

Kéo dài CK,DE cắt AB lần lượt ở M,N.
*Áp dụng hệ quả đ/l Thales, ta có:
 BMC có AE//CM (gt)  ABAM=BEECABAM=BEEC (1)
 BNE có BN//DC  BNDC=BEECBNDC=BEEC (2)
 AMK có AM//DC  AMDC=AKDKAMDC=AKDK (3)
*(1),(2) suy ra ABAM=BNDCABAM=BNDC
 ABBN=AMDCABBN=AMDC, lại có (3)
ABBN=AKDKABBN=AKDK 
 Q.E.D 

Answer 7

bài này khó qá

Answer 8

nếu làm bài này nhờ ai gigi mà l

Answer 9

Cam on