cho hình chữ nhật ABCD ( AB>BC) . Kẻ AH vuông góc với BD tại H 1) biết DH = 4,5cm, HB= 8cm. Tính AH, sinABD 2) đường thẳng AH cắt đoạn thẳng DC và đường thẳng BC lần lượt ở K và I . chứng minh tam giác DHK đồng dạng tam giác IHB và AH^2= HK.HI 3) qua điểm A kẻ đường thẳng song song với DB, đường thẳng này cắt đường thẳng BC tại điểm E. chứng minh 1/AB^2= 1/AI^2+ 1/BD^2 mng giúp e với ah :<< gấp lắm r ạ

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