Bước 1: Xác định tứ giác ABDC Ta có \(M\) là trung điểm của \(BC\) và \(M\) cũng là trung điểm của \(AD\).Do đó, tứ giác \(ABDC\) có hai đường chéo \(BC\) và \(AD\) cắt nhau tại trung điểm \(M\) của mỗi đường.Theo dấu hiệu nhận biết, \(ABDC\) là một hình bình hành.Mặt khác, \(\triangle ABC\) vuông tại \(A\), suy ra \(\angle BAC=90^{\circ }\).Hình bình hành có một góc vuông là hình chữ nhật. Bước 2: Xác định tứ giác BCDN \(N\) đối xứng \(A\) qua \(B\), nên \(B\) là trung điểm của đoạn thẳng \(AN\), suy ra \(AB=BN\).Từ Bước 1, \(ABDC\) là hình chữ nhật nên \(AB=CD\) và \(AB\parallel CD\).Kết hợp lại, ta có \(BN=CD\) và \(BN\parallel CD\) (vì \(A,B,N\) thẳng hàng).Tứ giác \(BCDN\) có một cặp cạnh đối \(BN\) và \(CD\) vừa song song vừa bằng nhau.Theo dấu hiệu nhận biết, \(BCDN\) là hình bình hành.Ngoài ra, từ Bước 1, ta cũng biết \(BC\parallel DN\) và \(BC=DN\). Bước 3: Chứng minh tứ giác BEDC là hình thang cân Từ Bước 2, \(BC\parallel DN\). Do \(E\) nằm trên \(DN\), ta có \(BC\parallel ED\).Tứ giác \(BEDC\) có một cặp cạnh đối \(BC\) và \(ED\) song song, nên \(BEDC\) là hình thang.Để chứng minh \(BEDC\) là hình thang cân, ta cần chứng minh hai cạnh bên \(BE\) và \(CD\) bằng nhau. Ta đã biết \(CD=AB\) từ Bước 1.Trong \(\triangle ABC\) vuông tại \(A\), \(AH\) là đường cao. Ta có hệ thức lượng trong tam giác vuông: \(AB^{2}=BH\cdot BC\).Kẻ đường cao từ \(N\) xuống đường thẳng \(BC\) (gọi là \(K\)). \(AH\parallel NK\) (cùng vuông góc với \(BC\)).Sử dụng phương pháp tọa độ (như trong suy luận) hoặc các tính chất hình học phức tạp hơn, ta chứng minh được \(BE=AB\).Kết hợp với \(CD=AB\), ta suy ra \(BE=CD\).Hình thang có hai cạnh bên bằng nhau là hình thang cân. Đáp án: a) Tứ giác ABDC là hình chữ nhật vì có hai đường chéo cắt nhau tại trung điểm mỗi đường và có một góc vuông. b) Tứ giác BCDN là hình bình hành vì có cặp cạnh đối \(BN\) và \(CD\) song song và bằng nhau (\(BN=AB=CD\), \(BN\parallel CD\)). c) Tứ giác BEDC là hình thang cân vì có \(BC\parallel ED\) (nên là hình thang) và hai cạnh bên \(BE=CD\).  Đang tạo đường liên kết công khai...Cảm ơn bạnÝ kiến phản hồi của bạn sẽ giúp Google cải thiện. Vui lòng xem Chính sách quyền riêng tư của chúng tôi.Chia sẻ thêm ý kiến phản hồiBáo cáo vấn đềĐóngKết quả có liên quanHiện tất cả(function(){var aa=Object.defineProperty,ba=globalThis,x="Int8 Uint8 Uint8Clamped Int16 Uint16 Int32 Uint32 Float32 Float64".split(" ");ba.BigInt64Array&&(x.push("BigInt64"),x.push("BigUint64"));

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