BDCN nội tiếp $\Rightarrow \widehat{BDN}=\widehat{BCN}$.Do đó $\widehat{ABN}=\widehat{BCM}\Rightarrow \Delta MNB\sim \Delta MBC\Rightarrow \frac{MB}{MC}=\frac{MN}{MB}\Rightarrow MB^{2}=MN.MC$
b,MA=MB $\Rightarrow MA^{2}=MN.MC\Rightarrow \Delta MNA\sim \Delta MAC \Rightarrow \widehat{MAN}=\widehat{MCA}$
$\widehat{MCA}=\widehat{NDC}\Rightarrow \widehat{MAN}=\widehat{NDC} \Rightarrow \widehat{MAD}=\widehat{ADC}\Rightarrow AB // CD$
c,ABCD là hình thoi $\left\{\begin{matrix} ABCD là hbh & & \\ AB=BC& & \end{matrix}\right.$
ABCD là hbh $\Rightarrow \widehat{BAC}=\widehat{BDC}$
$\widehat{ABC}=\widehat{BDC}\Rightarrow \widehat{BAC}=\widehat{ABC}\Rightarrow \Delta ABC$ đều =>$\widehat{BAC}=60^{\circ}$
ABCD là hình thoi => $\widehat{BAC}=\widehat{BDC}=60^{\circ}\Rightarrow \widehat{HDC}=60^{\circ} \Rightarrow \widehat{HCO}=60^{\circ}\Rightarrow HC=OC.cos30^{\circ}=R.\frac{\sqrt{3}}{2}\Rightarrow DH=HC.tan 30=\frac{R}{2}\Rightarrow Sabcd=4Sdhc=\frac{R^{2}.\sqrt{3}}{2}$
- onepiecekizaru yêu thích