16/11/2023 19:34:44

Cho hình bình hành ABCD, điểm G thuộc CD

cho hình bình hành ABCD ,điểm G thuộc CD.sao cho DG=1/4DC.Gọi E là giao điểm của AG và BD.TÍNH TỈ SỐ DE/DB

2 trả lời

+ Trả lời +3đ

Hỏi gia sư

Học gia sư

Tham khảo câu trả lời từ Trợ lý ảo Lazi

Để giải bài toán này, ta sẽ sử dụng định lí Menelaus.

Áp dụng định lí Menelaus cho tam giác ABD và đường chéo AG, ta có:
$\frac{DE}{DB} \cdot \frac{BG}{GA} \cdot \frac{AG}{AD} = 1$

Vì $BG = DC$ và $GA = GC - AC$, nên ta có:
$\frac{DE}{DB} \cdot \frac{DC}{GC - AC} \cdot \frac{AG}{AD} = 1$

Vì $GC = DC + DG = DC + \frac{1}{4}DC = \frac{5}{4}DC$, nên ta có:
$\frac{DE}{DB} \cdot \frac{DC}{\frac{5}{4}DC - AC} \cdot \frac{AG}{AD} = 1$

Vì $AC = AB$, nên ta có:
$\frac{DE}{DB} \cdot \frac{DC}{\frac{5}{4}DC - AB} \cdot \frac{AG}{AD} = 1$

Vì $AD = AB + BD$, nên ta có:
$\frac{DE}{DB} \cdot \frac{DC}{\frac{5}{4}DC - AB} \cdot \frac{AG}{AB + BD} = 1$

Vì $AG = AD - DG = AB + BD - \frac{1}{4}DC$, nên ta có:
$\frac{DE}{DB} \cdot \frac{DC}{\frac{5}{4}DC - AB} \cdot \frac{AB + BD - \frac{1}{4}DC}{AB + BD} = 1$

Rút gọn các thành phần trong phương trình trên, ta được:
$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4(AB + BD) - DC}{4(AB + BD)} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4(AB + BD) - DC}{4(AB + BD)} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4(AB + BD) - DC}{4(AB + BD)} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac{DE}{DB} \cdot \frac{4DC}{5DC - 4AB} \cdot \frac{4AB + 4BD - DC}{4AB + 4BD} = 1$

$\frac

Tham gia Cộng đồng Lazi trên các mạng xã hội
Fanpage:	https://www.fb.com/lazi.vn
Group:	https://www.fb.com/groups/lazi.vn
Kênh FB:	https://m.me/j/AbY8WMG2VhCvgIcB
LaziGo:	https://go.lazi.vn/join/lazigo
Discord:	https://discord.gg/4vkBe6wJuU
Youtube:	https://www.youtube.com/@lazi-vn
Tiktok:	https://www.tiktok.com/@lazi.vn

Vui	Buồn	Bình thường

Cho hình bình hành ABCD, điểm G thuộc CD

Đăng nhập

Mở khóa để xem toàn bộ nội dung trả lời