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Tính độ dài đoạn DN. Chứng minh DN vuông góc CE và CIN đồng dạng CBM

Cho hình vuông ABCD, gọi M,N lần lượt là trung điểm của cạnh AB,BC . Hai đường thẳng AD và MC cắt nhau tại E . Gọi I là giao điểm của CM và DN .a,cho độ dài cạnh vuông ABCD là 4cm . Tính độ dài đoạn DN. b, chứng minh DN vuông góc CE và CIN ~CBM . c, chứng minh :góc AIM + góc ACM =45°

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    Tham khảo câu trả lời từ Trợ lý ảo Lazi

    a, Ta có AM=MB=2cm, CN=NB=2cm.
    Gọi x là độ dài của DN, ta có:
    CM=CD+DM=4+2=6cm
    Theo định lí Menelaus, ta có:
    $\frac{CE}{EM}.\frac{MI}{ID}.\frac{DN}{NC}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
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    $\frac{CE}{2}.\frac{MI}{ID}.\frac{x}{2}=1$
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    $\frac{CE}{2}.\

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    0
    kí đầu màu giờ
    06/05/2024 21:00:17
    +5đ tặng
    Vì tứ giác ABCD là hình vuông 
    =>∠BCD = 90° 
    => ∆NDC vuông tại C
    Mà N là trung điểm BC 
    => NC = 2 cm
    Xét ∆NDC vuông tại C có:
    DN² = CN² + DC²
    DN² = 2²+4²
    DN = √20
    DN = 4,5 cm

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