mk ko biết cách vẽ hình trên olm nên bạn thông cảm

Vì d ko cắt BC => đường thẳng d // BC

=> \(\widehat{DAB}=\widehat{BAC},\widehat{DBC}=90^0\)

Xét tam giác ABC có \(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^0\)

                            => \(\widehat{ABC}+\widehat{ACB}=90^0\)

                          => \(\widehat{ABC}=90^0-\widehat{ACB}\)(1)

Ta lại có \(\widehat{DBC}=90^0\)=> \(\widehat{DAB}+\widehat{ABC}=90^0\)  

                                         => \(\widehat{ABC}=90^0-\widehat{DAB}\)(2)

Từ 1,2 => \(\widehat{ACB}=\widehat{DAB}\) 

mà \(\widehat{ABC}=\widehat{ACB}\)( Vì tam giác ABC cân tại A)

=> \(\widehat{DBA}=\widehat{ABC}\)

Mặt khác \(\widehat{DAB}=\widehat{ABC}\)(\(d//BC\))

=> \(\widehat{DAB}=\widehat{DBA}\)

=> tam giác DAB cân tại D => DA=DB

Tương tự :   AE=EC

=> BD + CE =AD+AE

=> BD+CE = DE (đpcm)

Ta có d đi qua A, D và E thuộc d 

=>D, A, E thẳng hàng  =>^DAB+^BAC+^CAE=180°  =>^DAB+^CAE=90°(1)

Xét tam giác DAB vuông ở D  =>^DBA+^DAB=90°(2) 

Từ (1) và (2)  =>^CAE=^DAB 

Xét tam giác BAD và tam giác ACE có:  ^DAB=^CAE(cmt) 

AB=AC(tam giác ABC cân)  ^ADB=^AEC(=90°) 

=>Tam giác BAD tam giác ACE(g.c.g)

=> BD=AE; EC=AD

Mà DE=AD+AE

=>DE=BD+CE

Làm ý a thôi!

a) \(\widehat{EAC}+\widehat{BAD}=180^o-90^o=90^o\)

Mà: \(\widehat{DBA}+\widehat{BAD}=90^o\)

=> \(\widehat{EAC}=\widehat{DBA}\)

Xét \(\Delta ABD\)và \(\Delta CAE\)

\(\hept{\begin{cases}AC=ABC\left(gt\right)\\\widehat{EAC}=\widehat{DBA}\\\widehat{EAC}=\widehat{BDA}\end{cases}}\)

\(\Rightarrow\Delta ABD=\Delta ACE\)

a) Ta có : BAD + BAC + CAE = 180 => BAD+CAE=90 (BAC=90)

mà CAE + ECA = 90 =>BAD=ECA

Xét tam giác BDA và tam giác AEC có: 

AC=AB (gt)

BAD=ECA

BDA=CEA=90 

=> tam giác BDA= tam giác AEC

b) =>AD=EC(t.ứng)

ta có: BD² + AD² = AB² hay BD² + EC² = AB²

a) Xét \(\Delta ABC\)và\(\Delta ADE\):

AB=AD(gt)

\(\widehat{BAC}=\widehat{DAE}=90^o\)

AC=AE(gt)

=> \(\Delta ABC=\Delta ADE\left(c-g-c\right)\)

=> BC=DE ( 2 cạnh tương ứng)

=> Đpcm

b) Ta có \(\Delta ABD\)vuông cân tại A

=> \(\widehat{ABD}=\widehat{ADB}=\frac{\widehat{DAB}}{2}=\frac{90^o}{2}=45^o\)

\(\Delta AEC\)vuông cân tại A

=> \(\widehat{AEC}=\widehat{ACE}=\frac{\widehat{EAC}}{2}=\frac{90^o}{2}=45^o\)

=> \(\widehat{BDA}=\widehat{ECA}=45^o\)

Mà 2 góc này ở vị trí so le trong

=> BD//CE

=> Đpcm

c) Sửa đề: Kẻ dường cao AH của tam giác ABC cắt DE tại M. Vẽ đường thẳng qua A và vuông góc với MC cắt BC tại N. Chứng minh rằng CA vuông góc với NM

Gọi giao điể của NA và MC là I

Xét \(\Delta NMC\)có:

\(\hept{\begin{cases}NI\perp MC\\MH\perp NC\end{cases}}\)

Mà 2 đường cao này cắt nhau tại A

=> A là trực tâm của \(\Delta MNC\)

=> \(CA\perp NM\)

=> Đpcm

d) Ta có: \(\widehat{ADM}=\widehat{ABC}\left(\Delta ADE=\Delta ABC\right)\)

=> \(\widehat{ADM}+\widehat{AED}=\widehat{ABC}+\widehat{BAH}=90^o\)

=> \(\widehat{AED}=\widehat{BAH}\) Mà \(\widehat{BAH}=\widehat{MAE}\left(đđ\right)\)

=> \(\widehat{AED}=\widehat{MAE}\)

=> \(\Delta MAE\)cân tại M

=> MA=ME (1)

Lại có: \(\widehat{AED}=\widehat{ACB}\Rightarrow\widehat{AED}+\widehat{ADE}=\widehat{ACB}+\widehat{CAH}=90^o\)

=> \(\widehat{ADE}=\widehat{CAH}\)

Mà \(\widehat{CAH}=\widehat{DAM}\left(đđ\right)\)

=> \(\widehat{ADE}=\widehat{DAM}\)

=> \(\Delta DAM\)cân tại M

=> MD=MA (2)

Từ (1) và (2)

=> MA=MD=ME

=> \(MA=\frac{1}{2}DE\)

=> Đpcm

P/s: Thật ra định làm tắt cho bạn tự suy luận, nhưng sợ bạn ko hiểu nên thoi, mỏi cả tay:>>>

a) Ta có : CE ⊥ d

                BD ⊥ d

\(\Rightarrow\)CE // BD  (ĐPCM)

b) Xét △CEA và △ADB có :

    AC = AB

   \(\widehat{EAC}=\widehat{ABD}\)(cùng phụ với \(\widehat{DAB}\))

\(\Rightarrow\) △CEA = △ADB (cạnh huyền-góc nhọn)

c) Có △CEA = △ADB

\(\Rightarrow\hept{\begin{cases}BD=AE\\CE=AD\end{cases}}\)(Cặp cạnh tương ứng)

\(\Rightarrow\)BD + CE = AE + AD = DE (ĐPCM)

d)  △ABC vuông tại A có AM là trung tuyến

\(\Rightarrow\)AM = BM = CM

\(\Rightarrow\)△ABM cân tại M

Có : \(\widehat{ECA}=\widehat{BAD}\)(△CEA = △ADB)

       \(\widehat{ACB}=\widehat{ABC}\) (△ABC cân tại A)

\(\Rightarrow\widehat{ECA}+\widehat{ACB}=\widehat{BAD}+\widehat{ABC}\)

Mà \(\widehat{ABC}=\widehat{MAB}\)(△MAC cân tại M)

\(\Rightarrow\widehat{ECA}+\widehat{ACB}=\widehat{BAD}+\widehat{MAB}\)

\(\Rightarrow\widehat{ECM}=\widehat{MAD}\)

Xét △ADM và △CEM có :

       EC = AD

       \(\widehat{ECM}=\widehat{MAD}\)

       AM = CM

\(\Rightarrow\)△ADM = △CEM (c-g-c)   (ĐPCM)

\(\Rightarrow\)EM = MD   (Cặp cạnh tương ứng) (1)

Có : \(\widehat{EMA}+\widehat{EMC}=90^o\)

       \(\widehat{EMC}=\widehat{DMA}\)(△ADM = △CEM)

\(\Rightarrow\widehat{EMA}+\widehat{DMA}=90^o\)

\(\Rightarrow\widehat{EMD}=90^o\)(2)

Từ (1) và (2) suy ra △DME vuông cân tại M.

mình không biết

Tài trợ

đề bài có thiếu ko bn?

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