Question

[happy new year]

Cho tam giác ABC Đều , E thuộc cạnh BC .Trên mp ko chứ A , dựng tai Bx sao cho góc CBx = góc CAE .Tia Bx cắt AE tại D

Chứng minh : DA=DB+DC

Answer 1

[Huyền Trân]

\(\text{Trên đoạn AD lấy E sao cho : BD=EF}\)

\(\Delta BED\text{có :}\)\(\widehat{EBD}+\widehat{BDE}+\widehat{DEB}=180^O\)

                 \(\Rightarrow\widehat{BDE}=180^O-\widehat{EBD}-\widehat{DEB}\)

                 \(\Rightarrow\widehat{BDE}=180^O-\widehat{CAE}-\widehat{AEC}\left(\widehat{DEB}=\widehat{AEC}\left(đ^2\right)\right)\)

                  \(\widehat{BDE}=\widehat{ACE}=60^O\)

          \(\text{Vì BD=EF}\)\(\widehat{BDF}=60^o\)

\(\Rightarrow\Delta BDF\text{ là tam giác đều}\)

\(\Rightarrow BD=BF=FD\)

\(\text{ta có :}\)\(\widehat{ABF}=60^O-\widehat{FBE}\)

\(\widehat{EBD}=60^o-\widehat{FBE}\)

\(\Rightarrow\widehat{ABF}=\widehat{EBD}\)

\(\text{Xét :}\)\(\Delta ABF\text{ và }\)\(\Delta CBD\text{ có}:\)

\(AB=AC\left(gt\right)\)

\(\widehat{ABF}=\widehat{EBD}\left(cmt\right)\)

\(BF=BD\left(cmt\right)\)

\(\Rightarrow\Delta ABF=\Delta CBD\left(c.g.c\right)\)

\(\Rightarrow AF=DC\)

\(\text{ta có : AF+FD=AD}\)

\(\Rightarrow DC+BD=AD\left(đpcm\right)\)