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hoangcuu
2 năm trước
Cho tam giác ADB vuông cân tại D (DA = DB) nội tiếp đường tròn tâm (O). Dựng hình bình hành ABCD. Gọi H là chân đường vuông góc kẻ từ D đến AC; K là giao điểm của AC với đường tròn (O). Chứng minh rằng: a) Tứ giác HBCD nội tiếp b) Góc DOK = 2* góc BDH c) CK*CA=2*BD2
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