\(\widehat{I_1}=\widehat{B_2}\)(2 góc slt của DE // BC) mà\(\widehat{B_1}=\widehat{B_2}\)(BI là phân giác góc ABC)\(\Rightarrow\widehat{I_1}=\widehat{B_1}\Rightarrow\Delta BDI\)cân tại D => BD = DI
\(\widehat{I_2}=\widehat{C_2}\)(2 góc slt của DE // BC) mà\(\widehat{C_1}=\widehat{C_2}\)(CI là phân giác góc ACB)\(\Rightarrow\widehat{I_2}=\widehat{C_1}\Rightarrow\Delta IEC\)cân tại E => IE = EC
Vậy DE = DI + IE = BD + CE (đpcm)
Vì DE song song với BC => \(\widehat{DIB}=\widehat{IBC}\) ( SLT) . Mà \(\widehat{IBC}=\widehat{DBI}\) ( BI là p/g của \(\widehat{ABC}\) ) => \(\widehat{DIB}=\widehat{DBI}\) theo định lý => tam giác DIB cân tại D => DB = DI
Vì DE song song với BC => \(\widehat{EIC}=\widehat{ICB}\)( SLT) .Mà \(\widehat{ECI}=\widehat{ICB}\) ( CI là p/g của \(\widehat{ECB}\) ) => \(\widehat{EIC}=\widehat{ECI}\) .Theo định lý => tam giác IEC cân tại E => EI = EC
Vì DE = DI + IE . Mà DI = DB ; IE = EC => DE = DB + CE
Vậy DE = DB + CE
Kẻ CI giao AB tại H, BI giao AC tại K
Ta có góc HIB = IBC + ICB (góc ngoài tam giác), DE // BC => HID = ICB
=> DIB = IBC mà BI là phân giác nên DBI = IBC => DIB = DBI => tam giác BDI cân tại D => DB = DI
Tương tự, ta có góc KIC = IBC + ICB (góc ngoài tam giác), DE // BC => KIE = IBC
=> EIC = ICB mà CI là phân giác nên ECI = ICB => EIC = ECI => tam giác CEI cân tại E => CE = EI
Ta có: DE = DI + IE mà DI = DB, IE = CE => DE = DB + CE => chứng minh được DE = BD + CE
vi DE song song BC suy ra DIB=IBC suy ra=DIB suy ra ID=DB
EIC=ICB =ICB IE=EC
suy ra DE=BD+CE
Cho tam giác ABC. Các tia phân giác của các góc B và C cắt nhau tại I. Chứng minh rằng AI là tia phân giác của góc A
Có DE\(//\)BC\(\Rightarrow\)\(\widehat{DIB}=\widehat{IBC}\)(2 góc so le trong)
Mà \(\widehat{IBC}=\widehat{IBD}\)\(\text{(BI phân giác của}\) \(\widehat{ABC}\))
\(\Rightarrow\widehat{DIB}=\widehat{IBD}\)\(\Rightarrow\Delta BID\)cân tại D\(\Rightarrow BD=DI\)
Có DE\(//\)BC\(\Rightarrow\)\(\widehat{EIC}=\widehat{ICB}\)(2 góc so le trong)
Mà \(\widehat{ICB}=\widehat{ICE}\)
\(\Rightarrow\widehat{EIC}=\widehat{ICE}\)\(\Rightarrow\Delta EIC\)\(\text{cân tại }\)E\(\Rightarrow EI=EC\)
\(\text{Lại có DE= DI+IE}\)\(\Rightarrow DE=BD+CE\)
\(\text{Có DE}\)\(//\)\(\text{BC}\Rightarrow\widehat{I_1}=\widehat{B_1}\)\(\text{(2 góc so le trong)}\)
\(\text{Mà }\widehat{B_1}=\widehat{B_2}\)(BI phân giác của \(\widehat{ABC}\))
\(\Rightarrow\widehat{I_1}=\widehat{B_2}\)\(\Rightarrow\Delta IBD\)cân tại D\(\Rightarrow BD=ID\)
có DE//BC\(\Rightarrow\)\(\widehat{C_1}=\widehat{I_2}\)(2 góc so le trong)
Mà \(\widehat{C_1}=\widehat{C_2}\)(CI phân giác của \(\widehat{ACB}\))
\(\Rightarrow\widehat{I_2}=\widehat{C_2}\)\(\Rightarrow\Delta ICE\)cân tại E\(\Rightarrow CE=IE\)
Mà DE=ID+IE\(\Rightarrow DE=BD+CE\)
Bạn Đinh Đức hùng làm đúng rồi đó bạn
~học tốt~
Bạn Đinh Đức Hùng làm đúng rồi đó bạn
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CIE = ICB (2 góc so le trong, DE // BC)
mà ICB = ICE (IC là tia phân giác của ECB)
=> CIE = ICE
=> Tam giác EIC cân tại I
=> EI = EC
BID = IBC (2 góc so le trong, DE // BC)
mà IBC = IBD (IB là tia phân giác của DBC)
=> BID = IBD
=> Tam giác DIB cân tại D
=> DI = DB
DE = DI + IE = DB + CE
