Giải:
Xét \(\Delta EAB,\Delta CAD\) có:
\(AE=AC\left(gt\right)\)
\(\widehat{EAB}=\widehat{CAD}\) ( đối đỉnh )
\(AB=AD\left(gt\right)\)
\(\Rightarrow\Delta EAB=\Delta CAD\left(c-g-c\right)\)
\(\Rightarrow\widehat{E_1}=\widehat{C_1}\) ( góc t/ứng )
\(\Rightarrow BE=CD\) ( cạnh t/ứng )
\(\Rightarrow\frac{1}{2}BE=\frac{1}{2}CE\)
\(\Rightarrow EM=NC\)
Xét \(\Delta MEA,\Delta NCA\) có:
\(EM=NC\left(cmt\right)\)
\(\widehat{E_1}=\widehat{C_1}\)
\(AE=AC\left(gt\right)\)
\(\Rightarrow\Delta MEA=\Delta NCA\left(c-g-c\right)\)
\(\Rightarrow AM=AN\) ( cạnh t/ứng )
\(\Rightarrowđpcm\)
Goi giao diem cua CF voi BE la M,giao diem cua EF voi CD la N.
theo t/c goc ngoai cua tam giac,ta co:
BMF=B+C1;BMF=F+E1
suy ra B+ C1=F+E1 (1)
tuong tu D+E2=F+C2 (2)
Theo gia thiet thi:C1=C2,E1=E2 (3)
Tu (1),(2) va (3),suy ra:
2F=B+D nen F=B+D/2
hay CFE=ABC+ACE/2
