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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)

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

13 tháng 2 2016

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7 tháng 3 2017

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

30 tháng 10 2017
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.
30 tháng 10 2017

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)

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

18 tháng 3 2018

cái thể loại 0 điểm hỏi đáp , đăng toán hình mà éo vẽ hình không = rác rưởi