cho tam giác ABC vuông cân tại A qua A vẽ đường thẳng d ở ngoài ABC. Vẽ BD\(\perp\) d, CE \(\perp\)d tại E, M là trung điểm BC. CMR:
a) BD +CE= DE
b) Tam giác MDE vuông cân
K
Khách
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CL
13 tháng 2 2016
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CV
7 tháng 3 2017
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
19 tháng 3 2017
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\(\Delta DAB\)và\(\Delta ECA\)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)
\(\Rightarrow\hept{\begin{cases}EC=AD\\BD=AE\end{cases}\Rightarrow BD+EC=AD+AE}=DE\)
TP
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.
TP
30 tháng 10 2017
tilado.edu.vn/student/facebook_view_question/code/747142 link đó bạn nào cần
16 tháng 2 2020
Tham khảo ở đây nha
https://olm.vn/hoi-dap/detail/12435070952.html
21 tháng 2 2020
Tham khảo ở đây nha
Câu hỏi của Phạm Hương Giang - Toán lớp 7 - Học toán với OnlineMath
cái thể loại 0 điểm hỏi đáp , đăng toán hình mà éo vẽ hình không = rác rưởi