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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
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
Có \(\hept{\begin{cases}\widehat{A_1}+\widehat{A_2}=90^o\\\widehat{A_1}+\widehat{B_1}=90^o\end{cases}\Rightarrow\widehat{A_2}=\widehat{B_1}}\)
Xét \(\Delta ADB\)và \(\Delta\)CEA có:
AB=AC (\(\Delta\)ABC cân tại A)
\(\widehat{A_2}=\widehat{B_1}\left(cmt\right)\)
\(\widehat{D}=\widehat{E}=90^o\)
=> \(\Delta ADB=\Delta CAE\left(ch-gn\right)\)
=> BD=AE
Ta có \(AE^2+CE^2=AC^2\)
=>\(BD^2+CE^2=AC^2\)
Vì AC không đổi => BD2+CE2 không đổi
Bài làm
Bài làm
Ta có: \(\widehat{DAB}+\widehat{BAE}=180^0\)( hai góc kề bù )
=> \(\widehat{DAB}+\widehat{BAC}+\widehat{CAE}=180^0\)
Hay \(\widehat{DAB}+90^0+\widehat{CAE}=180^0\)
=> \(\widehat{DAB}+\widehat{CAE}=180^0-90^0=90^0\) (1)
Xét tam giác ACE vuông ở E có:
\(\widehat{CAE}+\widehat{ECA}=90^0\) (2)
Từ (1), (2) => \(\widehat{ECA}=\widehat{DAB}\)
Lại xét tam giác ABD và tam giác CAE có:
\(\widehat{BDA}=\widehat{AEC}\left(=90^0\right)\)
Cạnh huyền AB = AC ( Do tam giác ABC vuông cân )
\(\widehat{ECA}=\widehat{DAB}\)( cmt )
Vậy tam giác ABD = tam giác CAE ( cạnh huyền - góc nhọn )
=> AD = EC ( hai cạnh tương ứng )
Xét tam giác ABD vuông ở D có:
AB2 = BD2 + AD2
Hay AB2 = BD2 + CE2
Mà AB luôn luôn không đổi.
=> Tổng của BD2 + CE2 có giá trị luôn không đổi/ ( đpcm )
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
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
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
tilado.edu.vn/student/facebook_view_question/code/747142 link đó bạn nào cần
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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
Tham khảo ở đây nha
https://olm.vn/hoi-dap/detail/12435070952.html
Tham khảo ở đây nha
Câu hỏi của Phạm Hương Giang - Toán lớp 7 - Học toán với OnlineMath