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DBAEC

xét △ABD có BD ⊥ AD nên vuông tại D

⇒ ^A1+^B1=900(1)

△ACE có CE ⊥ AE nên vuông tại E

⇒ ^A3+^C1=900(2)

^A2=900⇒^A1+^A3=180−^A2=900(3)

từ (1),(2),(3)⇒^A1=^C1

mà 2△ vuông ABD và ACE có cạnh huyền AB và AC bằng nhau (△ABC cân)

nên bằng nhau ⇒ AD = CE

AD2+BD2=AB2

⇔ CE2+BD2=AB2 không đổi

16 tháng 4 2018

xét △ABD có BD ⊥ AD nên vuông tại D

⇒ A1ˆ+B1ˆ=900(1)A1^+B1^=900(1)

△ACE có CE ⊥ AE nên vuông tại E

⇒ A3ˆ+C1ˆ=900(2)A3^+C1^=900(2)

A2ˆ=900⇒A1ˆ+A3ˆ=180−A2ˆ=900(3)A2^=900⇒A1^+A3^=180−A2^=900(3)

từ (1),(2),(3)⇒A1ˆ=C1ˆ(1),(2),(3)⇒A1^=C1^

mà 2△ vuông ABD và ACE có cạnh huyền AB và AC bằng nhau (△ABC cân)

nên bằng nhau ⇒ AD = CE

AD2+BD2=AB2AD2+BD2=AB2

⇔ CE2+BD2=AB2CE2+BD2=AB2 không đổi

DBAEC

xét △ABD có BD ⊥ AD nên vuông tại D

⇒ ^A1+^B1=900(1)

△ACE có CE ⊥ AE nên vuông tại E

⇒ ^A3+^C1=900(2)

^A2=900⇒^A1+^A3=180−^A2=900(3)

từ (1),(2),(3)⇒^A1=^C1

mà 2△ vuông ABD và ACE có cạnh huyền AB và AC bằng nhau (△ABC cân)

nên bằng nhau ⇒ AD = CE

AD2+BD2=AB2

⇔ CE2+BD2=AB2 không đổi

16 tháng 4 2018

xét △ABD có BD ⊥ AD nên vuông tại D

⇒ A1ˆ+B1ˆ=900(1)A1^+B1^=900(1)

△ACE có CE ⊥ AE nên vuông tại E

⇒ A3ˆ+C1ˆ=900(2)A3^+C1^=900(2)

A2ˆ=900⇒A1ˆ+A3ˆ=180−A2ˆ=900(3)A2^=900⇒A1^+A3^=180−A2^=900(3)

từ (1),(2),(3)⇒A1ˆ=C1ˆ(1),(2),(3)⇒A1^=C1^

mà 2△ vuông ABD và ACE có cạnh huyền AB và AC bằng nhau (△ABC cân)

nên bằng nhau ⇒ AD = CE

AD2+BD2=AB2AD2+BD2=AB2

⇔ CE2+BD2=AB2CE2+BD2=AB2 không đổi

13 tháng 2 2016

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7 tháng 3 2017

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

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)

10 tháng 11 2019

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

19 tháng 3 2017

A B C D E M d

a)  Ta có: \(\widehat{DAB}+\widehat{CAE}=180^0-\widehat{BAC}=90^0\)(1)

               \(\widehat{DAB}+\widehat{DBA}=180^0-\widehat{BDA}=90^0\)(2)

Từ (1) và (2) \(\widehat{DAB}+\widehat{CAE}=\widehat{DAB}+\widehat{DBA}\Rightarrow\widehat{CAE}=\widehat{DBA}\)

Xét\(\Delta DAB\)\(\Delta ECA\)có:\(\hept{\begin{cases}\widehat{BDA}=\widehat{AEC}=90^0\\AB=AC\\\widehat{DBA}=\widehat{CAE}\end{cases}\Rightarrow\Delta DAB=\Delta ECA}\)(cạnh huyền góc nhọn)

\(\Rightarrow\hept{\begin{cases}EC=AD\\BD=AE\end{cases}\Rightarrow BD+EC=AD+AE}=DE\)