Cho tam giác ABC có ba góc nhọn và AB < BC < AC nội tiếp đường tròn (O). Kẻ các đường cao BD, CE cắt nhau tại H (D ∈ AC, E ∈ AB). Chứng minh BCDE là tứ giác nội tiếp

cho Δ ABC có 3 góc nhọn, AB < BC < AC nội tiếp đường tròn (O). Kẻ các đường cao BD,      

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<!--[endif]-->                CE cắt nhau tại H (D ∈ AC, E ∈ AB)

a. c/m:  BCDE là tứ giác nội tiếp

    b. c/m: DA.DC = DH.DB

  c. Vẽ đường tròn tâm H, bán kính HA cắt các tia AB, AC lần lượt tại M, N. Chứng minh OA vuông góc với MN.

d. Các tiếp tuyến tại M , N của (H ; HA) cắt nhau tại P. Chứng minh AP đi qua trung điểm của BC.
hỏi ý c, d thôi ạ, ý a, b mình biết làm rồi ạ

----- Nội dung dịch tự động từ ảnh -----
Cho tam giác ABC có ba góc nhọn và AB < BC < AC nội tiếp
đường tròn (O). Kẻ các đường cao BD, CE cắt nhau tại H (D ∈
AC, E E AB).
1) Chứng minh BCDE là tứ giác nội tiếp.
2) Chứng minh DA.DC =DH.DB.
3) Vẽ đường tròn tâm H, bán kính HA cắt các tia AB, AC lần
lượt tại M, N. Chứng minh OA vuông góc với MN.
4) Các tiếp tuyến tại M, N của (H ; HA) cắt nhau tại P. Chứng
minh AP đi qua trung điểm của BC.