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

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

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

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

ΔΔ ADB vuông tại D nên: DBAˆ+DABˆ=900DBA^+DAB^=900 Lại có: EACˆ+DABˆ=1800−BACˆ=1800−900=900EAC^+DAB^=1800−BAC^=1800−900=900 ⇒⇒ DBAˆ=EACˆDBA^=EAC^ (1) ΔΔ ABC cân tại A nên AB = AC Kết hợp với (1) ⇒⇒ ΔADB=ΔCEAΔADB=ΔCEA (cạnh huyền - góc nhọn) ⇒BD=AE,AD=CE⇒BD=AE,AD=CE ⇒BD+CE=AE+AD=DE⇒BD+CE=AE+AD=DE b. ΔΔ AMB và ΔΔ AMC có: AB=ACAB=AC (ΔΔ ABC cân tại A) MB=MCMB=MC (M là trung điểm của BC) AM là cạnh chung ⇒ΔAMB=ΔAMC⇒ΔAMB=ΔAMC (c.c.c) ⇒MABˆ=MACˆ=900:2=450⇒MAB^=MAC^=900:2=450 Mà ΔΔ ABC vuông cân tại A nên: ABMˆ=450⇒MABˆ=ABMˆ=450ABM^=450⇒MAB^=ABM^=450 ⇒⇒ ΔΔ AMB vuông cân tại M ⇒⇒ MA=MBMA=MB Ta lại có: DBAˆ=EACˆ⇒DBAˆ+450=EACˆ+450DBA^=EAC^⇒DBA^+450=EAC^+450 ⇒DBAˆ+MBAˆ=EACˆ+MACˆ⇒MBDˆ=MAEˆ⇒DBA^+MBA^=EAC^+MAC^⇒MBD^=MAE^ Kết hợp với MA=MBMA=MB và BD=AEBD=AE ⇒⇒ ΔBDM=ΔAEMΔBDM=ΔAEM (c.g.c) ⇒BMDˆ=AMEˆ,MD=ME⇒BMD^=AME^,MD=ME (*) Lại có: DMAˆ+BMDˆ=DMAˆ+AMEˆ=900DMA^+BMD^=DMA^+AME^=900 (**) Từ (*) và (**) ta suy ra ΔΔ DME vuông cân tại M.

tilado.edu.vn/student/facebook_view_question/code/747142 link đó bạn nào cần

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

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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

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