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: Xét ΔBAE vuông tại A và ΔBDE vuông tại D co

BE chung

BA=BD

=>ΔBAE=ΔBDE

b: BA=BD

EA=ED

=>BE là trung trực của AD

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

BD=BA

góc B chung

=>ΔBDM=ΔBAC

=>BM=BC

=>ΔBMC cân tại B

`a,`

Xét `2 \Delta` vuông `ABE` và `DBE`:

`\text {BE chung}`

`\text {BA = BD (2 cạnh tương ứng)}`

`=> \Delta ABE = \Delta DBE (ch-cgv)`

`b,`

Gọi I là giao điểm của AD và BE

Vì `\Delta ABE = \Delta DBE (a)`

`->` $\widehat {ABE} = \widehat {DBE} (\text {2 góc tương ứng})$

Xét `\Delta ABI` và `\Delta DBI`:

`\text {BA = BD (gt)}`

$\widehat {ABI} = \widehat {DBI}$

`\text {BI chung}`

`=> \Delta ABI = \Delta DBI (c-g-c)`

`->` $\widehat {BIA} = \widehat {BID} (\text {2 cạnh tương ứng})$

Mà `2` góc này ở vị trí kề bù

`->` $\widehat {BIA} + \widehat {BID} = 180^0$

`->` $\widehat {BIA} = \widehat {BID} =$\(\dfrac{180}{2}=90^0\)

`-> \text {BI} \bot \text {AD}` 

Mà `\text {I} \in \text {BE}`

`-> \text {BE} \bot \text{AD}`

`c,`

Vì `\Delta ABE = \Delta DBE (a)`

`-> \text {AE = DE (2 cạnh tương ứng)}`

Xét `\Delta AEM` và `\Delta DEC`:

`\text {AE = DE}`

$\widehat {AEM} = \widehat {DEC} (\text {2 góc đối đỉnh})$

$\widehat {MAE} = \widehat {CDE} (=90^0)$

`=> \Delta AEM = \Delta DEC (cgv-gn)`

`-> \text {AM = DC (2 cạnh tương ứng)}`

Ta có: \(\left\{{}\begin{matrix}\text{BM = AM + AB}\\\text{BC = BD + DC}\end{matrix}\right.\)

Mà \(\left\{{}\begin{matrix}\text{BA = BD}\\\text{AM = DC}\end{matrix}\right.\)

`-> \text {BM = BC}`

Xét `\Delta MBC`:

`\text {BM = BC}`

`-> \Delta MBC` cân tại B.

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) Áp dụng định lí Pi - ta - go cho tam giác ABC vuông tại A có :

AB^2+AC^2 =BC^2hay AC^2=15^2-9^2=144 hay AC=12

b)Xét tam giác ABE và DBE có :

     Góc A=góc B(=90 độ)

     BA=BD(gt)

     Chung cạnh BE

suy ra tam giác ABE= BDE (c.g.c)

c) Từ tam giác ABE=BDE(cm ở ý b) suy ra góc ABE = góc DBE (2 góc tương ứng )

            Suy ra BE là tia phân giác cua góc ABC

Xét tam giác BDK và BAC có :

       Chung góc B

       BA=BD(gt)

       góc D = góc A (=90 độ)

suy ra tam giác BDK=tam giác BAC (g.c.g)

suy ra AC=DK (2 cạnh tương ứng ) 

                  ( Mình chỉ làm được ý a,b,c thôi , mình ngại vẽ hình . Nếu đúng kết bạn với mình nhé )

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

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

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

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

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

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

\(BE\text{ chung}\)

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

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

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

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

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

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

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

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

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

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

\(BO\text{ chung}\)

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

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

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

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

\(\text{Mà chúng kề bù}\)

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

\(\Rightarrow BE\perp AD\)

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

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

a) 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)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