Cho đường tròn (O) và điểm A nằm ngoài đường tròn đó. Từ điểm A kẻ hai tiếp tuyến AM, AN với đường tròn (O) (M,N là 2 tiếp điểm ). Kẻ cát tuyến ABC không đi qua O (B nằm giữa A và C ) gọi H là trung điểm của BC
a) chứng minh rằng các điểm o,h,m,a,n cùng thuộc 1 đường tròn
b) chứng minh HA là tia phân giác góc MHN
Từ điểm A nằm ngoài đường tròn (O;R) vẽ các tiếp tuyến AB, AC với đường tròn (B, C là các tiếp điểm). Trên bán kính OC lấy điểm M. Tia AM cắt (O) tại D và E (D nằm giữa A và E). Đoạn thẳng OA cắt BC tại H.
a) Chứng minh 4 điểm A, B, O, C cùng thuộc một đường tròn
b) Chứng minh AC2=AD.AE.
c) Chứng minh góc AHD = góc AEO
d) Vẽ đường thẳng qua O vuông góc với DE và vẽ tiếp tuyến của đường tròn (O) tại E. Hai đường thẳng này cắt nhau tại I. Chứng minh B, C, I thẳng hàng.
a) Xét (O):
AB là tiếp tuyến; B là tiếp điểm (gt). \(\Rightarrow\widehat{ABO}=90^o.\)
AC là tiếp tuyến; C là tiếp điểm (gt). \(\Rightarrow\widehat{ACO}=90^o.\)
\(\Rightarrow\) 4 điểm A, B, O, C cùng thuộc một đường tròn đường kính AO.
b) Xét (O):
\(\widehat{ACD}=\widehat{AEC}\) (Góc tạo bởi tia tiếp tuyến và dây; góc nội tiếp cùng chắn \(\stackrel\frown{CD}\)).
Xét \(\Delta ACD\) và \(\Delta AEC:\)
\(\widehat{ACD}=\widehat{AEC}\left(cmt\right).\)
\(\widehat{CAD}chung.\)
\(\Rightarrow\Delta ACD=\Delta AEC\left(g-g\right).\)
\(\Rightarrow\dfrac{AC}{AE}=\dfrac{AD}{AC}.\\ \Rightarrow AC^2=AD.AE.\)
Bài 5: (3,0 điểm). Cho tam giác ABC nhọn nội tiếp đừng tròn tâm O. Các đường cao AD, BE và CF cắt
nhau tại H. Đường thẳng EF cắt đường tròn ở I và K. Chứng minh:
a) Tứ giác BCEF nội tiếp.
b) AE.AC = AF.AB.
c) AI = AK.
câu C the nao ạ? cám ơn mọi nguoi
a Xét tứ giác BCEF có
\(\widehat{BFC}=\widehat{BEC}=90^0\)
Do đó:BCEF là tứ giác nội tiếp
b: Xét ΔABE vuông tại E và ΔACF vuông tại F có
\(\widehat{BAE}\) chung
DO đó: ΔABE\(\sim\)ΔACF
Suy ra: AB/AC=AE/AF
hay \(AB\cdot AF=AE\cdot AC\)
Cho hai đường tròn (O) và (O’) cắt nhau tại A và B. Một đường thẳng điểm đi qua B cắt (O) ở M và (O’) ở N (M và N khác B). Các tiếp tuyến tại M và N của hai đường tròn cắt nhau ở P. a) Tính MPN cho biết MAN . b) Chứng tỏ rằng: MNP vuông tại P o OAO 90 .
Câu 4(3d) Cho đường tròn tâm O, đường kính AB. Lấy điểm C trên đường tròn sao cho số đo c AC bằng 60°. Vẽ dây CD vuông góc với AB tại M, dây BC cắt đường tròn đường kí BO tại điểm thứ hai là I a) Chứng minh Ol//AC và ba điểm O:I;D thẳng hàng b)Xác định vị trí tương đối của đường thẳng MI với đường tròn đường kính BO c) Gọi N là điểm nằm giữa O và D. Tia CN cắt đường thẳng AD tại K, đường thẳng cắt đường thẳng OK tại E. Chứng minh tứ giác AOED nội tiếp
Cho nửa đường tròn (O;R) đường kính AB,H là trung điểm AO kẻ đường thẳng qua H vuông góc AB ,cắt nửa đường (O;R) tại C .Kẻ tia phân giác của góc BAC cắt (O) tại D ,cắt BC tại K
Cho ba điểm A, B, C thẳng hàng (B nằm giữa A và C) vẽ đường tròn tâm O đường kính BC; AM là tiếp tuyến vẽ từ A (M là tiếp điểm). Từ M vẽ dây MN vuông góc với BC tại H. a) Chứng minh MB là tia phân giác của góc AMN; b) Từ B vẽ đường thẳng song song với MC, đường thẳng này cắt MN, MA lần lượt tại D và E. Chứng minh: AB.HC = AC. HB
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Cho tam giác ABC nhọn, có AB<AC. vẽ đường cao AD, đường phân giác AO của tam giác ABC, vẽ (O) tiếp xúc với AB,AC lần lượt ở M,N. a)cm:M,N,O,D,A cùng thuộc 1 đ tròn. b)CM: góc BMD =góc CDN. c) qua O kẻ đường thẳng vuông góc với BC cắt MN ở I. AI cắt BC ở K. cm: K là trung điểm của BC.
Cho (O) có đường kính AB=2R, C là trung điểm của OA và dây MN vuông góc với OA tại C. Gọi E là điểm tùy ý trên cung nhỏ BM, H là giao điểm của AE và MN CMR: MA là tiếp tuyến của đường tròn ngoại tiếp tam giác AMH.
bt:giải hệ phương trình sau
a,\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
Đặt \(\dfrac{1}{x-y+2}=a;\dfrac{1}{x+y-1}=b\)
Ta có HPT
\(\left\{{}\begin{matrix}14a-10b=9\\3a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14a-10b=9\\15a+10b=20\end{matrix}\right.\Leftrightarrow}}\left\{{}\begin{matrix}29a=29\\3a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=\dfrac{1}{2}\end{matrix}\right.\)
