Question

[Phạm Quốc Dân]

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

Answer 1

[ღᏠᎮღ🅼🅸🅽🅷ঔ🅽🅶ụ🆈ঌ__[ᴅʀᴇᴀм тᴇᴀм]__]

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 \(\widehat{DBC}=\widehat{DCB}+\widehat{BDC}\left(dpcm\right)\)

Answer 2

[ღᏠᎮღ🅼🅸🅽🅷ঔ🅽🅶ụ🆈ঌ__[ᴅʀᴇᴀм тᴇᴀм]__]

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)