Em xem lại đề nhé \(\frac{\widehat{ABC}+\widehat{ADE}}{2}\)

EM là phân giác của góc DEA

=>\(\hat{MEA}=\frac12\cdot\hat{DEA}\)

CM là phân giác của góc ACB

=>\(\hat{MCA}=\frac12\cdot\hat{ACB}\)

\(\hat{MEC}+\hat{MCE}\)

\(=\hat{MEA}+\hat{MCA}+\hat{AEC}+\hat{ACE}\)

\(=\frac12\cdot\left(\hat{AED}+\hat{ACB}\right)+180^0-\hat{EAC}=\frac12\left(\hat{AED}+\hat{ACB}\right)+\hat{BAC}\)

\(=\frac12\left(\hat{AED}+\hat{ACB}\right)+\hat{BAC}\)

\(=\frac12\cdot\left(180^0-\hat{DAE}-\hat{ADE}+\hat{ACB}\right)+\hat{BAC}=90^0-\frac12\cdot\hat{BAC}-\frac12\cdot\hat{ADE}+\hat{BAC}+\frac12\cdot\hat{ACB}\)

\(=90^0+\frac12\cdot\hat{BAC}-\frac12\cdot\hat{ADE}+\frac12\cdot\hat{ACB}=90^0+\frac12\cdot\left(180^0-\hat{ABC}\right)-\frac12\cdot\hat{ADE}\)

\(=180^0-\frac12\cdot\left(\hat{ABC}+\hat{ADE}\right)\)

XétΔEMC có \(\hat{EMC}+\hat{ECM}+\hat{MEC}=180^0\)

=>\(\hat{EMC}=180^0-180^0+\frac12\cdot\left(\hat{ABC}+\hat{ADE}\right)\)

\(=\frac12\cdot\left(\hat{ABC}+\hat{ADE}\right)\)

=>ĐPCM

Kẻ OF//BC(F thuộc AC)

=>OF//DE//BC

DE//BC

=>góc DEA=góc ACB

=>góc DEO=1/2*góc ACB

ED//OF
=>góc DEA=góc CFD và góc DEO=góc EOF

=>góc EOF=1/2*góc ACB

=>góc DEO=góc EOF

OF//BC

=>góc FOB=góc OBC=1/2góc ABC

góc BOE=góc BOF+góc EOF

=1/2(góc ABC+góc ACB)

\(a.\) Ta có: \(\widehat{B}=2\widehat{C}\)suy ra \(\widehat{C}=\frac{\widehat{B}}{2}\)                                                    \(\left(1\right)\)
Vì \(BD\)là tia phân giác của \(\widehat{B}\)suy ra \(\widehat{ABD}=\widehat{DBC}=\frac{\widehat{B}}{2}\)                \(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra \(\widehat{ABD}=\widehat{DBC}=\widehat{C}\)
- Xét \(\Delta ABD\)có     \(\widehat{ADB}+\widehat{DBA}+\widehat{BAD}=180^0\)(đ/lý tồng 3 góc trong cùng 1 tam giác)
                         \(\Rightarrow\)\(\widehat{ADB}+\widehat{BAD}=180^0-\widehat{DBA}\)
- Xét \(\Delta ABC\)có       \(\widehat{BAC}+\widehat{ACB}+\widehat{CBA}=180^0\)
                         \(\Rightarrow\) \(\widehat{BAC}+\widehat{CBA}=180^0-\widehat{ACB}\)
        mà  \(\widehat{ACB}=\widehat{ABD}\)(cmt)     suy ra  \(\widehat{BAC}+\widehat{CBA}=\widehat{ADB}+\widehat{BAD}\)

- Xet  \(\Delta ABD\)có  \(\widehat{ABE}\)là góc ngoài tại đỉnh \(B\)
                     suy ra  \(\widehat{ABE}=\widehat{ADB}+\widehat{BAD}\) 
- Xet  \(\Delta ABC\)có  \(\widehat{ACK}\)là góc ngoài tại đỉnh \(C\)
                     suy ra  \(\widehat{ACK}=\widehat{ABC}+\widehat{BAC}\) 
    mà    \(\widehat{BAC}+\widehat{CBA}=\widehat{ADB}+\widehat{BAD}\)        \(\Rightarrow\)đpcm

\(b.\)  Xét  \(\Delta AEB\)và  \(\Delta KCA\) có:     \(AB=CK\)         ( gt )
                                                             \(\widehat{ABE}=\widehat{ACK}\)      ( cmt )
                                                                \(EB=AC\)          ( gt )
                   Do đó  \(\Delta AEB\)\(=\)\(\Delta KCA\) (c.g.c)