Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
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\(\Delta DAB\)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)
\(\Rightarrow\hept{\begin{cases}EC=AD\\BD=AE\end{cases}\Rightarrow BD+EC=AD+AE}=DE\)
cái thể loại 0 điểm hỏi đáp , đăng toán hình mà éo vẽ hình không = rác rưởi
A B C D E M N H
a) Xét \(\Delta ABC\)và\(\Delta ADE\):
AB=AD(gt)
\(\widehat{BAC}=\widehat{DAE}=90^o\)
AC=AE(gt)
=> \(\Delta ABC=\Delta ADE\left(c-g-c\right)\)
=> BC=DE ( 2 cạnh tương ứng)
=> Đpcm
b) Ta có \(\Delta ABD\)vuông cân tại A
=> \(\widehat{ABD}=\widehat{ADB}=\frac{\widehat{DAB}}{2}=\frac{90^o}{2}=45^o\)
\(\Delta AEC\)vuông cân tại A
=> \(\widehat{AEC}=\widehat{ACE}=\frac{\widehat{EAC}}{2}=\frac{90^o}{2}=45^o\)
=> \(\widehat{BDA}=\widehat{ECA}=45^o\)
Mà 2 góc này ở vị trí so le trong
=> BD//CE
=> Đpcm
c) Sửa đề: Kẻ dường cao AH của tam giác ABC cắt DE tại M. Vẽ đường thẳng qua A và vuông góc với MC cắt BC tại N. Chứng minh rằng CA vuông góc với NM
Gọi giao điể của NA và MC là I
Xét \(\Delta NMC\)có:
\(\hept{\begin{cases}NI\perp MC\\MH\perp NC\end{cases}}\)
Mà 2 đường cao này cắt nhau tại A
=> A là trực tâm của \(\Delta MNC\)
=> \(CA\perp NM\)
=> Đpcm
d) Ta có: \(\widehat{ADM}=\widehat{ABC}\left(\Delta ADE=\Delta ABC\right)\)
=> \(\widehat{ADM}+\widehat{AED}=\widehat{ABC}+\widehat{BAH}=90^o\)
=> \(\widehat{AED}=\widehat{BAH}\) Mà \(\widehat{BAH}=\widehat{MAE}\left(đđ\right)\)
=> \(\widehat{AED}=\widehat{MAE}\)
=> \(\Delta MAE\)cân tại M
=> MA=ME (1)
Lại có: \(\widehat{AED}=\widehat{ACB}\Rightarrow\widehat{AED}+\widehat{ADE}=\widehat{ACB}+\widehat{CAH}=90^o\)
=> \(\widehat{ADE}=\widehat{CAH}\)
Mà \(\widehat{CAH}=\widehat{DAM}\left(đđ\right)\)
=> \(\widehat{ADE}=\widehat{DAM}\)
=> \(\Delta DAM\)cân tại M
=> MD=MA (2)
Từ (1) và (2)
=> MA=MD=ME
=> \(MA=\frac{1}{2}DE\)
=> Đpcm
P/s: Thật ra định làm tắt cho bạn tự suy luận, nhưng sợ bạn ko hiểu nên thoi, mỏi cả tay:>>>