\(\text{a)Xét }\Delta ABC\text{ vuông tại A có:}\)

\(BC^2=AB+AC^2\left(\text{định lí Py ta go}\right)\)

\(\Rightarrow BC^2=5^2+7^2=25+49=74\left(cm\right)\)

\(\Rightarrow BC=\sqrt{74}\left(cm\right)\)

\(\text{b)Xét }\Delta ABE\text{ và }\Delta DBE\text{ có:}\)

\(\widehat{BAE}=\widehat{BDE}=90^0\left(gt\right)\)

\(BE\text{ chung}\)

\(BA=BD\left(gt\right)\)

\(\Rightarrow\Delta ABE=\Delta DBE\left(c-g-c\right)\)

\(\text{c)Xét }\Delta AEF\text{ và }\Delta DEC\text{ có:}\)

\(\widehat{AEF}=\widehat{DEC}\left(\text{đối đỉnh}\right)\)

\(\widehat{FAE}=\widehat{CDE}=90^0\left(gt\right)\)

\(AE=DE\left(\Delta ABE=\Delta DBE\right)\)

\(\Rightarrow\Delta AEF=\Delta DEC\left(g-c-g\right)\)

\(\Rightarrow EF=EC\left(\text{hai cạnh tương ứng}\right)\)

\(\text{d)Gọi O là giao điểm của BE và AD}\)

\(\text{Xét }\Delta ABO\text{ và }\Delta DBO\text{ có:}\)

\(BO\text{ chung}\)

\(BA=BD\left(gt\right)\)

\(\widehat{ABO}=\widehat{DBO}\left(\Delta ABE=\Delta DBE\right)\)

\(\Rightarrow\Delta ABO=\Delta DBO\left(c-g-c\right)\)

\(\Rightarrow\widehat{AOB}=\widehat{DOB}\left(\text{hai góc tương ứng}\right)\)

\(\text{Mà chúng kề bù}\)

\(\Rightarrow\widehat{AOB}=\widehat{DOB}=\dfrac{180^0}{2}=90^0\)

\(\Rightarrow BE\perp AD\)

\(\text{Mà AO=DO}\left(\Delta AOB=\Delta DOB\right)\)

\(\Rightarrow BE\text{ là đường trung trực của đoạn thẳng AD}\)

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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a) Áp dụng định lí pytago cho \(\Delta ABC\):

\(BC^2=AB^2+AC^2\)

\(BC^2=5^2+7^2\)

\(BC^2=25+49\)

\(BC^2=74\)

\(BC=\sqrt{74}\)

\(\Rightarrow BC=\sqrt{74}\)

Chúc bn hk tốt :D

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

a) dùng pyta go

b) = nhau theo trường hợp cạnh huyền cạnh góc vuông

c) dựa vào kết quả câu b =>tam giác AEF=tam giác DEC

d)tam giác ABD cân có BE là phân giác =>đpcm

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

để suy nghĩ xem đã!!!!!!!!!!

câu này khó ak!!!

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

a: Xét ΔBAE vuông tại A và ΔBDE vuông tại D có

BE chung

BA=BD

=>ΔBAE=ΔBDE

b: ΔBAE=ΔBDE

=>AE=DE

c: Xét ΔBDF vuông tại D và ΔBAC vuông tại A có

BD=BA

góc B chung

=>ΔBDF=ΔBAC

=>BF=BC

=>ΔBFC cân tại B

mà BN là trung tuyến

nên BN là phân giác của góc FBC

mà BE là phân giác của góc ABE

nên B,E,N thẳng hàng

a) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

BC²=AB²+AC²

⇔BC²=3²+4²=25=5²

sorry bt mỗi câu a hoi

ok nha đợi minh một lát

câu b/ Xét tg ABD và tg EBD có:

BD cạnh chung

ABD=EBD ( do BD là tia phân giác ABC)

BAD=BED (=90)

=> tg ABD= tg EBD (cạnh huyền_góc nhọn)