Xét tứ giác $ABNC$ có:
$AM = MN$
$BM = CM$
$\widehat{A} = 90^o$
Do đó $ABNC$ là hình chữ nhật
$\Rightarrow \widehat{CBN} =\widehat{ANB} =\widehat{ACB}$
a) Ta có:
$BC.\sin^2\widehat{CBN}$
$= BC.\sin^2\widehat{ACB}$
$= BC.\dfrac{AB^2}{BC^2}$
$= \dfrac{AB^2}{BC} = BH$
b) $\dfrac{1}{2\sin^2\widehat{ANB}}$
$= \dfrac{1}{2\sin^2\widehat{ACB}}$
$= \dfrac{1}{2\dfrac{AB^2}{BC^2}}$
$= \dfrac{BC}{2}\cdot\dfrac{1}{\dfrac{AB^2}{BC}}$
$= \dfrac{BM}{HB}$
$= \dfrac{HB + HM}{HB}$
$= \dfrac{HM}{HB} + 1$
