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Hương Giang
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N.T.M.D
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N.T.M.D
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Nhok_baobinh
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Phúc
1 tháng 12 2017 lúc 12:01

2 truong hop nhu nhau ma.

TH1 neu AE=CH,BE=AH

Ap dung dinh li py ta go ta co

Do AEB la tam giac vuong

=> AB2=AE2+BE2(1)

Do AHC la tam giac vuong

=> AC2=AH2+HC2(2)

Ma AE=CH,BE=AH(3) 

Từ 1 2 3 => AB=AC

Th 2: AE=AH,BE=CH lam tt

Phúc
1 tháng 12 2017 lúc 12:06

Bạn tự cm tứ giác AEBF và tứ giác AHCK là hcn nhe 

dsfdsf
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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:31

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Mèo Chó
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nguyễn quỳnh trang
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nguyễn quỳnh trang
13 tháng 11 2016 lúc 19:59
mọi người ơi giúp mình với.
nguyễn lan anh
13 tháng 11 2016 lúc 20:29

2.tự vẽ hình nhe

xét tam giác abc có

Góc CAx= góc B+góc C =40 + 10=80<đlí góc ngoài tam giác>

Vì Ac là phân giác của A

Góc A1=A2=1/2A=40

Ta có A2=C=40

Mà hai góc này ở vị trí so le trong

suy ra ax song song BC

Hồ Phúc huynh
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41 Võ Minh Quân
12 tháng 1 2022 lúc 13:10

ΔABC có : A^+ABC^+C^=1800 ( tổng ba góc của một tam giác )

⇒600+ABC^+700=1800

⇒ABC^=1800−(700+600)=500

Mà ABC^+ABD^=1800 ( hai góc kề bù )

⇒500+ABD^=1800

41 Võ Minh Quân
12 tháng 1 2022 lúc 13:12

⇒ABD^=1800−500=1300

Vậy 

Tươi Lưu
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Rhider
31 tháng 1 2022 lúc 8:52

undefined

a) Xét   \(\Delta ABC\) có tia phân giác \(BAC,ACB\)  cắt nhau tại O suy ra O là giao điểm của 3 đường phân giác trong tam giác ABC suy ra BO là phân giác của \(\widehat{CBA}\)   (tính chất 3 đường phân giác của tam giác)

\(\Rightarrow DBO=ABO=\dfrac{DBA}{2}\left(1\right)\) ( tính chất tia phân giác )

Lại có BF là phân giác của \(\widehat{ABx\left(gt\right)}\) \(=ABF=FBx\left(2\right)\)

( tính chất của tia phân giác ) 

Mà \(ABD+ABx=180^o\left(3\right)\left(kềbu\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow OBA+ABF=180^o\div2=90^o\Rightarrow BO\text{⊥ }BF\)

b) Ta có \(FAB+BAC=180^o\)( kề bù ) mà \(BAC=120^o\left(gt\right)\Rightarrow FAB=60^o\)

\(\Rightarrow\text{AD là phân giác của}\widehat{BAC}\)  ( dấu hiệu nhận biết tia phân  giác )

\(\Rightarrow BAD=CAD=60^o\) ( tính chất tia phân giác )

\(\Rightarrow FAy=CAD=60^o\) ( đối đỉnh ) \(\Rightarrow FAB=FAy=60^o\Rightarrow\) AF là tia phân giác của \(BAy\) ( dấu hiệu nhận biết tia phân giác )

Vậy \(\Delta ABD\) có hai tia phân giác của hai góc ngoài tại đỉnh A và đỉnh B cắt nhau tại F nên suy ra DF là phân giác của \(ADB=BDF=ADF\) ( tính chất tia phân giác )

c) Xét \(\Delta ACD\) có phân giác góc ngoài tại đỉnh A và phân giác trong tại đỉnh C cắt nhau tại E nên suy ra DE cũng là phân giác của \(ADB\Rightarrow\)\(D,E,F\) thẳng hàng