$\boxed{18}$Cho tứ giác $ABCD$. Biết rằng $AB.CD=AD.BC$. Chứng minh rằng: $\widehat{ABD}+\widehat{ACB}=\widehat{ACD}+\widehat{ADB}$
Lời giải bài $\boxed{18}$:
Trên nửa mặt phẳng bờ $AD$ không chứa $C$ dựng điểm $E$ sao cho $\widehat{DAE}=\widehat{BAC},\widehat{ADE}=\widehat{ABC}$
$\Rightarrow \Delta ADE\sim\Delta ABC(g.g)\Rightarrow \frac{AD}{AB}=\frac{AE}{AC},\widehat{AED}=\widehat{ACB}$
Suy ra $\Delta ACE\sim\Delta ABD(c.g.c)\Rightarrow \widehat{ACE}=\widehat{ABD},\widehat{AEC}=\widehat{ADB}$
Ta có: $\frac{DE}{BC}=\frac{AD}{AB}=\frac{CD}{BC}\Rightarrow DE=CD\Rightarrow \widehat{DCE}=\widehat{DEC}$
Vậy $\widehat{ABD}+\widehat{ACB}=\widehat{ACE}+\widehat{AED}=\widehat{ACD}-\widehat{DCE}+\widehat{AEC}+\widehat{DEC}=\widehat{ACD}+\widehat{AEC}=\widehat{ACD}+\widehat{ADB}$