Bai 1Cho tam giac ABC can o A tren tia doi AC lay AD=AC
1) tam giac ABD la tam giac gi
2) chung minh goc DBC=go BDC+go DCB
3) tinh goc DBC
Bai 2 Cho goc xOy<90 do,lay A,B thuo Ox(A nam giua O va B),lay C,D thuoc Oy sao cho OA=OC,AB=CD
1) Chung minh tam giac OBD an
2) So sanh AD va BC
3) Goi I la giao diem AD va BC tam giac IBD va tam giac IAC la cac tam giac gi
4)cm tam giac OAI=tam giac OCI
Bai 3 cho tam giac ABC can tai A ,lay diem D thuoc AB.Trn tia doi cua tia CA lay CE=BD,DE cat BC o M
1)Chung minh M la trung diem DE
Bai 4 cho tam giac ABC nhon o A =60 do ,hai duong phan giac BD va CE cat nhau tai Ii
1)Tinh BIC
2)IE la duong phan giac cua tam giac IBC
Chung minh+)tam giac BIE=tam giac BIF
+)tam giac CID=tam giac CIF
1.a) \(\Delta ABC\)cân tại A\(\Rightarrow AB=AC\).Mà \(AD=AC\Rightarrow AB=AD\)
Xét \(\Delta ABD\)có \(AB=AD\Rightarrow\Delta ABD\)cân tại A
b)Có \(\widehat{ABC}=\widehat{ACB}\left(1\right)\)( do \(\Delta ABC\)cân)
\(\widehat{ABD}=\widehat{ADB}\left(2\right)\)( do \(\Delta ABD\)cân )
Từ \(\left(1\right);\left(2\right)\Rightarrow\widehat{ABC}+\widehat{ABD}=\widehat{ACB}+\widehat{ADB}\)
\(\Rightarrow\widehat{DBC}=\widehat{ACB}+\widehat{ADB}\)hay \(\widehat{DBC}=\widehat{DCB}+\widehat{BDC}\left(dpcm\right)\)
2.
a)Nối A vs C
có\(OA=0C;AB=CD\Rightarrow OA+AB=OC+CD\)
hay \(OB=OD\).Xét \(\Delta OBD\)có \(OB=OD\Rightarrow\Delta OBD\)cân tại O
b) Xét \(\Delta OAD\)và \(\Delta OCB\)có:
\(OA=OB\left(gt\right)\)
\(\widehat{AOB}:chung\)
\(OB=OD\left(cmt\right)\)
\(\Rightarrow\Delta OAD=\Delta OCB\left(c.g.c\right)\Rightarrow AD=CB\left(dpcm\right)\)
c)Có \(\Delta OAD=\Delta OCB\Rightarrow\widehat{ADO}=\widehat{CBO}\)
Xét \(\Delta ACD\)và \(\Delta CBA\)có: \(AD=CD\)
\(\widehat{ADO}=\widehat{CBO}\)
\(CD=BA\)
\(\Rightarrow\Delta ACD=\Delta CBA\left(c.g.c\right)\Rightarrow\widehat{CAD}=\widehat{BCA}\Rightarrow\Delta IAC\)cân tại I
Làm tương tự bạn => tam giác IBD cân tại I ( tam giác ADB = tam giác CBD => Góc ADB= góc CBD)