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Jkkgng
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Võ Mỹ Hảo
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Mon an
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Nguyễn Lê Phước Thịnh
3 tháng 12 2023 lúc 21:06

a: Xét ΔABH có BI là phân giác

nên \(\dfrac{AI}{AB}=\dfrac{IH}{BH}\)

Xét ΔABC có BD là phân giác

nên \(\dfrac{AD}{AB}=\dfrac{CD}{CB}\)

Đề bài này chưa đủ dữ kiện để tính cụ thể AI/AB; AD/AB nha bạn

b: ΔBAD vuông tại A

=>\(\widehat{ABD}+\widehat{ADB}=90^0\)

=>\(\widehat{ADI}+\dfrac{1}{2}\cdot\widehat{ABC}=90^0\left(1\right)\)

ΔBIH vuông tại H

=>\(\widehat{HBI}+\widehat{BIH}=90^0\)

=>\(\widehat{BIH}+\dfrac{1}{2}\cdot\widehat{ABC}=90^0\)(2)

Từ (1) và (2) suy ra \(\widehat{ADI}=\widehat{BIH}\)

mà \(\widehat{AID}=\widehat{BIH}\)(hai góc đối đỉnh)

nên \(\widehat{ADI}=\widehat{AID}\)

=>ΔAID cân tại A

=>AD=AI(3)

Xét ΔABH có BI là phân giác

nên \(\dfrac{IH}{BH}=\dfrac{AI}{AB}\left(4\right)\)

Xét ΔABC có BD là phân giác

nên \(\dfrac{DC}{BC}=\dfrac{DA}{AB}\left(5\right)\)

Từ (3),(4),(5) suy ra \(\dfrac{IH}{BH}=\dfrac{DC}{BC}\)

Khánh
10 tháng 12 2023 lúc 19:18

1+1=2

kkkkk
13 giờ trước (20:58)

 

Lê Quỳnh Hương
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Lê Quỳnh Hương
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Kim San
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Võ Mỹ Hảo
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Đặng An Na
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%Hz@
13 tháng 6 2020 lúc 15:41

A)XÉT \(\Delta ABD\)\(\Delta HBD\)

\(\widehat{BAD}=\widehat{BHD}=90^o\)

\(\widehat{ABD}=\widehat{DBH}\left(GT\right)\)

BD LÀ CẠNH CHUNG

=>\(\Delta ABD\)=\(\Delta HBD\)(CẠNH HUYỀN - GÓC NHỌN ) ( ĐPCM)

GỌI I LÀ GIAO ĐIỂM CỦA BD VÀ AH

XÉT \(\Delta ABI\)\(\Delta HBI\)

\(AB=BH\left(\Delta ABD=\Delta HBD\right)\)

\(\widehat{ABD}=\widehat{DBH}\left(GT\right)\)

BI LÀ CẠNH CHUNG

=>\(\Delta ABI\)=\(\Delta HBI\)(C-G-C)

\(\Rightarrow\widehat{AIB}=\widehat{HIB}\)( HAI GÓC TƯƠNG ỨNG)

MÀ HAI GÓC NÀY KỀ BÙ 

\(\Rightarrow\widehat{AIB}=\widehat{HIB}=\frac{180^o}{2}=90^o\left(1\right)\)

\(\Delta ABI\)=\(\Delta HBI\)(C-G-C)

=> AI=HI( HAI CẠNH TƯƠNG ỨNG ) (2)

TỪ 1 VÀ 2 => BI LÀ ĐƯỜNG TRUNG TRỰC CỦA AH HAY BD LÀ ĐƯỜNG TRUNG TRỰC CỦA AH(ĐPCM)

B)

b)  

Vì  \(\Delta\)DBA =\(\Delta\) DBH ( cm ở câu a )

=) AD = DH 

Xét\(\Delta\)DHC ( DHC = 90 ) có :

DC là cạnh huyền 

\(\Rightarrow\) DC là cạnh lớn nhất 

\(\Rightarrow DC>DH\)

mà DH = AD

\(\Rightarrow AD< DC\)

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Nhật Hạ
13 tháng 6 2020 lúc 15:31

a, Xét △ABD vuông tại A và △HBD vuông tại H

Có: BD là cạnh chung

       ABD = HBD (gt)

=> △ABD = △HBD (ch-gn)

=> AB = BH (2 cạnh tương ứng) => B thuộc đường trung trực của AH

và AD = HD (2 cạnh tương ứng) => D thuộc đường trung trực của AH

=> BD là đường trung trực của AH

b, Xét △HDC vuông tại H có: DC > DH (quan hệ giữa đường xiên và đường vuông góc)

=> DC > AD

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Greninja
13 tháng 6 2020 lúc 15:50

a) Xét \(\Delta ABD\)và \(\Delta HBD\)có :

                  \(\widehat{BAD}=\widehat{AHD}\left(=90^o\right)\)

                \(BD\)chung

                  \(\widehat{B_1}=\widehat{B_2}\left(gt\right)\)

\(\Rightarrow\Delta ABD=\Delta HBD\left(ch-gn\right)\)

\(\Rightarrow AB=BH\)( 2 cạnh tương ứng ) \(\Rightarrow\)B thuộc đường trung trực của AH \(\left(1\right)\)

và \(AD=HD\)( 2 cạnh tương ứng ) \(\Rightarrow\)D thuộc đường trung trực của AH \(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\)BD là trung trực của AH

b) Xét \(\Delta DHC\)vuông tại H , ta có :

      \(DH< DC\left(cgv< ch\right)\)

mà \(AD=HD\left(cmt\right)\)

\(\Rightarrow AD< DC\)

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Trần Ngự Thư
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Hoàng Trang
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Cao Linh Chi
13 tháng 2 2016 lúc 11:30

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CUTE vô đối
7 tháng 3 2017 lúc 20:37

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

cô nàng bạch dương
18 tháng 3 2017 lúc 12:04

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