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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
Tự vẽ hình nha bạn
Ta có: tam giác ABC cân tại A
=> B = C
Ta có: Góc D = góc E = 90o (góc vuông)
K1 = K2 (2 góc đối đỉnh)
=> 180 - E - K1 = 180 - D - K2
=> B1 = C1
Vì B = C ; B1 = C1 => B - B1 = C - C1
=> B2 = C2
Vì B2 = C2 nên KBC cân tại K
=> KB = KC
Xét tam giác AKB và tam giác AKC có:
AK cạnh chung (1)
AB = AC (2)
BK = BC (3)
Từ (1) ; (2) ; (3) = > Tam giác AKB = tam giác AKC (c - c - c) (4)
Từ (4) = > A1 = A2 (2 góc tương ứng)
=> AK là tia phân giác của góc A
=> ĐPCM
Tớ sẽ bổ sung thêm hình sau
a) Xét tam giác vuông ADB và tam giác vuông ACE có:
Góc A chung
AB = AC (gt)
\(\Rightarrow\Delta ABD=\Delta ACE\) (Cạnh huyền - góc nhọn)
b) Do \(\Delta ABD=\Delta ACE\Rightarrow AD=AE\)
Xét tam giác vuông AEH và tam giác vuông ADH có:
Cạnh AH chung
AE = AD (cmt)
\(\Rightarrow\Delta AEH=\Delta ADH\) (Cạnh huyền - cạnh góc vuông)
\(\Rightarrow HE=HD\)
c) Xét tam giác ABC có BD, CE là đường cao nên chúng đồng quy tại trực tâm. Vậy H là trực tâm giác giác.
Lại có AM cũng là đường cao nên AM đi qua H.
d) Xét các tam giác vuông EBC và EAC, áp dụng định lý Pi-ta-go ta có:
\(BC^2=EB^2+EA^2;AC^2=EA^2+EC^2\)
Tam giác ABC cân tại A nên AB = AC hay \(AB^2=AC^2\)
Vậy nên \(AB^2+AC^2+BC^2=2AC^2+BC^2=2\left(EA^2+EC^2\right)+EB^2+EC^2\)
\(=3EC^2+2EA^2+BC^2\).
A) Xét tam giác BEC và tam giác CDB có :
\(\widehat{BEC}\)=\(\widehat{CDB}\)=\(90^0\)
\(BC\)chung
\(\widehat{EBC}\)=\(\widehat{DCB}\)( giả thiết )
\(\Rightarrow\Delta EBC=\Delta DCB\left(G-C-G\right)\)
Vậy \(BD=CE\) ( hai canh tương ứng )
B) Xét tam giác DHC và tam giác EHC có :
\(\widehat{EBH}\) =\(\widehat{DCH}\)( vì góc CDH=góc BEB ; góc EHB = góc DHC )
EB=DC ( theo phần a )
\(\widehat{HEB}\)=\(\widehat{CDH}\)=900
\(\Rightarrow\)\(\Delta EHB=\Delta DHC\left(G-C-G\right)\)
\(\Rightarrow BB=HC\)( HAI CẠNH TƯƠNG ỨNG )
\(\Rightarrow\Delta BHC\)cân ( định lí tam giác cân )
C) Ta có : AB =AC ( giả thiêt )
Vậy góc A cách đều hai mút B và C
Vậy AH là đường trung trực của BC
d)Xét tam giác BDC và tam giác KDC có :
DK=DB ( GT )
CD ( chung )
suy ra tam giác BDC =tam giác KDC ( cạnh huyền - cạnh góc vuông )
\(\Rightarrow\) \(\widehat{BCD}\)=\(\widehat{KCD}\)( HAI GÓC TƯƠNG ỨNG )
Mà ta lai có góc EBC = góc BCD theo giả thiết )
\(\Rightarrow\)\(\widehat{EBC}\)=\(\widehat{EBC}\)
chúc bạn hok giỏi
Câu hỏi của Marklin_9301 - Toán lớp 8 - Học toán với OnlineMath
Em tham khảo tại đây nhé.
c)
Ta thấy EB = AE
Mà theo quan hệ giữa đường vuông góc với đường xiên thì AC < AE
Vậy nên AC < EB.