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![](https://rs.olm.vn/images/avt/0.png?1311)
hình tự vẽ:
xét hai tam giác vuông ABE và DBE:
ab=ad(gt); be là cạnh huyền chung
=>\(\Delta\) ABE = \(\Delta\)DBE
mình sẽ giải tiếp
a) theo đinh j lý pitago : tam giác abc vuông tại A
=> \(AB^2+AC^2=BC^2\)THAY SỐ TA ĐƯỢC \(5^2+7^2=BC^2\) TA ĐƯỢC \(74=BC^2\) =>BC =
8.6023
![](https://rs.olm.vn/images/avt/0.png?1311)
a, xét tam giác ABC theo định lý py _ta _go ta có :
\(^{BC^2=AC^2+AB^2}\)
\(BC^2=5^2+7^2\)
\(^{BC^2=25+49}\)
\(^{BC^2=74}\)
BC=\(\sqrt{74}\)
b,xét tam giác vuông ABE và tam giác vuông DBE ta có:
BA=DB(gt)
BE chung
=}tam giác ABE=tam giác DBE(ch_cgv)
=}EA=ED (2 cạnh tương ứng)
c,xét tam giác vuông AEF và tam giác vuông DEC ta có:
AE=ED(cm câu b)
E1=E2 (đối đỉnh)
=}tam giác AEF và tam giác DEC (gn_cgv)
=}EF=EC (2 cạnh tương ứng)
d,Ta có :BA =DA (gt)
AE=ED(cm câu a)
=}BE là đường trung trực của AD
MÌNH TỰ LÀM KHÔNG BIẾT CÓ ĐÚNG HAY KHÔNG BẠN Ạ
a) Xét tam giác ABC vuông tại A
có: \(AB^2+AC^2=BC^2\) ( py - ta - go )
thay số: \(5^2+7^2=BC^2\)
\(BC^2=74\)
\(\Rightarrow BC=\sqrt{74}\)cm
b) Xét tam giác ABE vuông tại A và tam giác DBE vuông tại D
có: AB = DB ( gt)
AE là cạnh chung
\(\Rightarrow\Delta ABE=\Delta DBE\left(ch-cgv\right)\)
c) ta có: tam giác ABE = tam giác DBE ( phần b)
=> AE = DE ( 2 cạnh tương ứng)
Xét tam giác AEF vuông tại A và tam giác DEC vuông tại D
có: AE = DE ( cmt)
góc AEF = góc DEC ( đối đỉnh )
\(\Rightarrow\Delta AEF=\Delta DEC\left(cgv-gn\right)\)
=> EF = EC ( 2 cạnh tương ứng)
d) ta có: tam giác ABE = tam giác DBE ( phần b)
=> góc ABE = góc DBE ( 2 góc tương ứng )
Xét tam giác ABH và tam giác DBH
có: AB = DB ( gt)
góc ABE = góc DBE ( cmt)
BH là cạnh chung
\(\Rightarrow\Delta ABH=\Delta DBH\left(c-g-c\right)\)
=> AH = DH ( 2 cạnh tương ứng ) (1)
góc AHB = góc DHB ( 2 góc tương ứng )
mà góc AHB + góc DHB = 180 độ ( kề bù)
=> góc AHB + góc AHB = 180 độ
2. góc AHB = 180 độ
góc AHB = 180 độ :2
góc AHB = 90 độ
=> \(\Rightarrow BE\perp AD⋮H\) ( định lí vuông góc) (2)
Từ (1) ; (2) => BE là đường trung trực của AD ( định lí đường trung trực)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Vì tam giác BAC vuông tại A
=> AB^2 + AC^2 = BC^2 ( đl pytago )
=> BC^2 = 5^2 + 7^2 = 74
=> BC = căn bậc 2 của 74
b)
Xét tam giác ABE; tam giác DBE có :
AB = DB ( gt)
góc ABE = góc DBE ( gt)
BE chung
=> tam giác ABE = tam giác DBE (c.g.c) - đpcm
c)
Vì tam giác ABE = tam giác DBE (câu b)
=> AE = DE
Xét tg AEF ⊥ tại A; tg DEC ⊥ tại D:
AE = DE (c/m trên)
g AEF = g DEC (đối đỉnh)
=> tg AEF = tg DEC (cgv - gn) - đpcm
=> EF = EC
d)
Do tam giác AEF = tam giác DEC (câu c)
=> AE = DE
=> E ∈ đường trung trực của AD (1)
Lại do AB = BD (gt)
=> B ∈ đường trung trực của AD (2)
Từ (1) và (2) => BE là đường trung trực của AD. - đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Áp dụng pytago .
b) Xét t/g ABE; tg DBE:
AB = DB ( gt)
g ABE = DBE (suy từ gt)
BE chung
=> tg ABE = tg DBE (c.g.c)
c) Vì tg ABE = tg DBE (câu b)
=> AE = DE
Xét tg AEF ⊥⊥ tại A; tg DEC ⊥⊥ tại D:
AE = DE (c/m trên)
g AEF = g DEC (đối đỉnh)
=> tg AEF = tg DEC (cgv - gn)
=> EF = EC
d) Do tg AEF = tg DEC (câu c)
=> AE = DE
=> E ∈∈ đg trung trực của AD (1)
Lại do AB = BD (gt)
=> B ∈ đg trung trực của AD (2)
Từ (1) và (2) => BE là đg trung trực của AD.
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) tam giác ABC vuông tại A
=> AB2 + AC2 = BC2
=> 52 + 72 = BC2
=> BC2 = 25 + 49 = 74
=> BC = \(\sqrt{74}cm\)
hình như bn ghi sai đề rùi làm sao làm bài b) !!!!!!!1
7756
![](https://rs.olm.vn/images/avt/0.png?1311)
a)tg BAC vuông tại A suy ra AB^2+AC^2=BC^2(định lý pi-ta-go)
suy ra BC^2=5^2+7^2=74
suy ra BC=\(\sqrt{74}\)
b)tg ABE=tgDBE(ch cgv)suy ra AE=ED
c)tg AEF=DEC(g c g) suy ra EF=EC(2 cạnh tương ứng )
d)gọi I là giao điểm của AD và BE
ta có AB=BD suy ra tgABD cân tại B
tg ABE=DBE(cmt) suy ra góc ABE=DBE mà BE nằm giữa 2 tia AB và BD suy ra BE là tia phân giác của góc ABD
tg cân ABD có BI là tia phân giác của góc ABD suy ra BI còn là đường trung trực của AD suy ra BE là đường trung trực của AD
Bạn tự vẽ hình nhá
a.Vì \(\Delta ABC\perp A\). Theo định lí Pytago ta có
\(BC^2\)= \(AB^2\)+\(AC^2\)
\(BC^2\)= \(5^2+7^2\)
\(BC^2\)= 74
Vì BC >.0 => BC = \(\sqrt{74}\)cm
b.Xét 2 tam giác vuông ABE và DBE có
AB = BD (Gt)
BE chung
=> \(\Delta ABE=\Delta DBE\) ( cạnh huyền-cạnh góc vuông )
c. Xét 2 tam giác vuông AEF và DEC có
AE = ED ( \(\Delta ABE=\Delta DBE\) )
\(\widehat{AEF}\)=\(\widehat{DEC}\)( đđ )
=> \(\Delta AEF=\Delta DEC\)(cạnh góc vuông-góc nhọn kề cạnh đó)
=> EF = EC ( t.ứ )
d. Ta có : AB = BD => B nằm trên đường trung trực của AD (1)
Vì \(\Delta AEF=\Delta DEC\) => EA = ED
=> E nằm trên đường trung trục của AD ( 2)
Từ (1) và (2) => BE là đường trung trực của AD
Chúc bạn học giỏi !
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