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Toán Hình THCS
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Aug.21
5 tháng 6 2019 lúc 12:25

bạn tự vẽ hình nhé !

                                                                    Giải

a,Ta có :\(\widehat{BAB'}=\widehat{AB'A'}=\widehat{B'A'B}=1v\)( nội tiếp nửa đường tròn )

\(\Rightarrow ABA'B'\)là hình chữ nhật

b, Ta có : BH // CA' (cùng vuông góc với AC )

               BA' // CH ( cùng vuông góc với AB )

\(\Rightarrow BHCA'\)là hình bình hành nên BH = CA' 

 c, \(\Delta BHC=\Delta BA'C\)nên đường tròn ngoại tiếp tam giác BHC bằng đường tròn ngoại tiếp tam giác BA'C

Mà đường tròn ngoại tiếp tam giác BA'C chính là đường tròn (O)

Vậy bán kính đường tròn ngoại tiếp tam giác BHC bằng R

 a) tứ giác ABA'B' có AA', BB' là hai đương chéo bằng nhau ( = 2R) 
=> ABA'B' là hình chữ nhật. 

b) ta có : 
CH _I_ AB ( H là trực tâm của tam giác ABC ) 
A'B _I_ AB ( ABA' chắn nửa đường tròn ) 
=> CH // A'B (1) 
Lại có : 
BH _I_ AC ( H là trực tâm của tam giác ABC ) 
A'C _I_ AC ( ACA' chắn nửa đường tròn ) 
=> A'C // BH (2) 
(1),(2) => BHCA' là hình bình hành 
=> BH=CA' 

c) kéo dài AH cắt đường tròn ngoại tiếp ABC tại D. Dễ dàng nhận thấy D và H đối xứng nhau qua BC ---> tam giác BCD = tam giác BCH --> đường tròn ngoại tiếp BCH = đường tròn ngoại tiếp BCD (đồng thời ngoại tiếp ABC) --> bán kính đường tròn ngoại tiếp BHC = R 

Toán Hình THCS
11 tháng 6 2019 lúc 10:37

cảm ơn bạn nha !

Nguyễn Thị Thảo
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linh nguyễn
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Quỳnh Trang
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dsfdsf
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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:31

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Văn Phúc Đạt lớp 9/7 Ngu...
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Hồng Nhan
10 tháng 4 2022 lúc 21:13

Xét tứ giác AEDB có: \(\widehat{AEB} = \widehat{ADB} = 90^o \)

⇒ Tứ giác AEDB nội tiếp (2 đỉnh E và D kề nhau cùng nhìn AB dưới 1 cặp góc bằng nhau)

⇒ \(\widehat{EAD} = \widehat{EBD} \) (cùng chắn \(\stackrel\frown{\text{ED}}\))

Xét ΔADC và ΔHDB có:

\(\widehat{ADC} = \widehat{HDB} = 90^o\)

\(\widehat{CAD} = \widehat{HBD} \)   (cmt)

⇒ ΔADC ∼ ΔBDH (g-g)

Văn Phúc Đạt lớp 9/7 Ngu...
10 tháng 4 2022 lúc 21:12

giúp mik vs

 

Trần Hoàng Minh
Xem chi tiết
Nguyễn Lê Phước Thịnh
9 tháng 10 2021 lúc 21:10

a: Xét tứ giác AMCN có 

AM//CN

AM=CN

Do đó: AMCN là hình bình hành

Suy ra: Hai đường chéo AC và MN cắt nhau tại trung điểm của mỗi đưòng

hay OA=OC; OM=ON

Nguyễn Tuấn Anh
Xem chi tiết
Nguyễn Tuấn Anh
1 tháng 5 2021 lúc 10:20

Bạn nào lướt qua thì giúp mình phần c với nha :v hơi bí phần c

ljiijl
1 tháng 5 2021 lúc 12:56

chứng minh cho DE sog sog vs A'C = cách cm 2 góc SLT ∠EDC=∠DCA'

đến đó tự lm i

pansak9
Xem chi tiết
Nguyễn Thảo Trang
7 tháng 11 2021 lúc 16:57

đường thẳng m cắt AC và BD tại A và B

mà m đi qua 2 đường thẳng và tạo thành 2 cặp góc vuông

⇒AC//BD

Nguyễn Thảo Trang
7 tháng 11 2021 lúc 16:59

-vì AC//BD và P cắt AC và BD tại C và D

-theo định luật 1 đường thẳng cắt hai đường thẳng thì sẽ có cặp góc đồng vị bằng nhau

⇒ C=D (=70o)

Pham Trong Bach
Xem chi tiết
Cao Minh Tâm
22 tháng 2 2019 lúc 8:15

= (1 + √a)(1 - √a)

= 1 - (√a)2 = 1 - a = VP (đpcm)