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5 tháng 5 2015

a và b. Xét tam giác ABD và ACE

 (chung)

AB = AC

Suy ra tam giác ABD = tam giác ACE ---> AE = AD 

Vậy tam giác AED là tam giác cân.

c)Xin lỗi nha mình không giải được

d) Ta có CD vuông góc với BK. vậy CD là đường cao của tam giác CBK mà BD = DK do đó đường cao trùng với đường trung trực. Suy ra tam giác cân ---> DKC = DBC

Mà góc ACE = ABD. Vậy suy ra góc ECB = DBC mà DBC = DKC --> ECB = DKC.

5 tháng 5 2015

ukm cũng cảm ơn bạn                                                  

8 tháng 5 2016

a) Xét tg ABD và tg ACE có

A là góc chung

E = D = 90 độ

AB = AC ( do tg ABC cân tại A )

=> tg ABD = tg ACE ( cạnh huyền - góc nhọn )

b) Vì tg ABD = tg ACE (cmt) => AD = AE ( 2 cạnh tương ứng )

Có : AE + EB = AB ; AD + DC = AC

mà AB = AC ( cmt ) ; AD = AE ( cmt )

=> EB = DC

Xét tg EBC và tg DCB có :

E = D = 90 độ

B = C ( do tg ABC cân )

EB = DC (cmt)

=> tg EBC = tg DCB (gcg)

=>

10 tháng 5 2016

không có câu c) à

13 tháng 1 2019

chị làm đây ko bt đúng hay sai đâu nha

xét tam giác ABC có BD vuông góc với AC

                               CE vuông góc với AB 

                               hai đường thẳng này cát nhau tại I 

suy ra I là trực tâm của tam giác ABC

suy ra AI vuông góc với BC(1)

Mặt khác, M là trung điểm của BC=> AM là đường trung tuyến của tam giác ABC

mà trong 1 tam giác cân đường trung tuyến đồng thời là đường cao

<=> AM cũng là đường cao của tam giác ABC

=> AM vuông góc với BC(2)

từ (1)(2) ta có A,I,M thẳng hàng

13 tháng 2 2016

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7 tháng 3 2017

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a: Xét ΔADB vuông tại D và ΔACE vuông tại E có

AB=AC

góc BAD chung

=>ΔADB=ΔACE

b: Xét ΔIBC có góc IBC=góc ICB

nên ΔIBC cân tại I

23 tháng 4 2017

A) Xét tam giác BEC và tam giác CDB có :

            \(\widehat{BEC}\)=\(\widehat{CDB}\)=\(90^0\)

          \(BC\)chung

          \(\widehat{EBC}\)=\(\widehat{DCB}\)( giả thiết )

       \(\Rightarrow\Delta EBC=\Delta DCB\left(G-C-G\right)\)

       Vậy \(BD=CE\)   ( hai canh tương ứng )

B) Xét tam giác DHC và tam giác EHC có :

         \(\widehat{EBH}\)  =\(\widehat{DCH}\)( vì góc CDH=góc BEB ; góc EHB = góc DHC )

          EB=DC ( theo phần a )

         \(\widehat{HEB}\)=\(\widehat{CDH}\)=900

            \(\Rightarrow\)\(\Delta EHB=\Delta DHC\left(G-C-G\right)\)

       \(\Rightarrow BB=HC\)( HAI CẠNH TƯƠNG ỨNG )

\(\Rightarrow\Delta BHC\)cân ( định lí tam giác cân )

         C) Ta có : AB =AC ( giả thiêt )

     Vậy góc A cách đều hai mút B và C 

       Vậy AH là đường trung trực của BC

   d)Xét tam giác BDC và tam giác KDC có : 

 DK=DB ( GT )

     CD ( chung )

     suy ra tam giác BDC =tam giác KDC ( cạnh huyền - cạnh góc vuông )

    \(\Rightarrow\) \(\widehat{BCD}\)=\(\widehat{KCD}\)( HAI GÓC TƯƠNG ỨNG ) 

   Mà ta lai có góc EBC = góc BCD  theo giả thiết )

         \(\Rightarrow\)\(\widehat{EBC}\)=\(\widehat{EBC}\)

  chúc bạn hok giỏi 

17 tháng 6 2022

ủa bạn hình như câu d 2 Tgiac=nhau theo TH 2cgv mà bạn