Đáp án:
Giải thích các bước giải:
Ta có: $\widehat{A_3}=\widehat{A_1}=180^o-\widehat{A_2}$
$\widehat F+\widehat {A_3}+\widehat{D_1}=180^o$
$\Rightarrow\widehat F+\widehat {A_3}+(\widehat E+\widehat C)=180^o$
$\Rightarrow \widehat F+\widehat E=180^o-\widehat {A_3}-\widehat C$
$=180^o-(180^o-\widehat{A_2})-\widehat C$
$=\widehat {A_2}-\widehat C$
$\Rightarrow\dfrac{\widehat F+\widehat E}{2}=\widehat {F_2}+\widehat{E_2}=\dfrac{\widehat{A_2}-\widehat C}2$(1)
$\widehat{M_1}=\widehat{E_2}+\widehat{G_1}$
$=\widehat{E_2}+(\widehat{F_2}+\widehat C)$
Thay (1) vào phương trình ta có:
$\widehat {M_1}=\dfrac{1}{2}(\widehat{A_2}-\widehat C)+\widehat C$
$=\dfrac{1}{2}(\widehat{A_2}+\widehat C)$
$=\dfrac{1}{2}.180^o$ (do tứ giác $ABCD$ có các góc đối bù nhau)
$=90^o$
$\Rightarrow FM\bot EM$ (đpcm).
