Cho hình thang cân ABCD (AD//BC) đường cao BE cắt đường chéo AC tại F,đường thẳng AB và CD cắt nhau tại M. Biết AB=20cm, AF/FC=2/3. Tính BM?

By Trang Ánh Nam -

Cho hình thang cân ABCD (AD//BC) đường cao BE cắt đường chéo AC tại F,đường thẳng AB và CD cắt nhau tại M. Biết AB=20cm, AF/FC=2/3. Tính BM?

 

 

Giải: dựng MH là đường cao của tam giác ADM.

Bài này T sẽ giải theo kiểu của lớp 8 vì T thấy sử dụng đồng dạng khá nhiều. Nhưng tôi không chứng minh 2 tam giác đồng dạng mà chỉ ghi ra thôi, các bạn tự chứng minh nếu cần thiết.

Ta có: tam giác AEF đồng dạng với CBF

Lại có: tam giác ABE đồng dạng với AMH

Đáp án; BM=15 cm.

 

Comments

  1. Trang Ánh NamFebruary 12, 2019 at 10:01 AM

    ở chổ "Lại có" các bạn có thể dùng Thales cũng được, nếu bạn nhớ.

    còn đây là 1 cách giải khác:

    Ta có: AF/FC =AE/BC =2/3

    Do ABCD là h.t cân => AD = 2AE +BC

    => AD/BC = (2AE + BC)/BC = 2AE/BC + 1 = 4/3 + 1 =7/3

    Mà AD/BC = AM/BM

    => AM/BM = 7/3

    => (AB+BM)/BM = AB/BM + 1 =7/3

    => AB/BM =4/3

    => BM = 3AB/4 = 15

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