Xét $ΔABH$ vuông tại $H$
$sinB=\dfrac{AH}{AB}$
$⇒AB=\dfrac{AH}{sinB}$
$⇒AB=\dfrac{2}{sin45^o}$
$⇒AB=2\sqrt[]{2}cm$
$cosB=\dfrac{BH}{AB}$
$⇒BH=AB.cosB$
$⇒BH=2\sqrt[]{2}.cos45^o$
$⇒BH=2cm$
Xét $ΔABC$ vuông tại $A$
$\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}$
$⇒\dfrac{1}{2^2}=\dfrac{1}{(2\sqrt[]{2})^2}+\dfrac{1}{AC^2}$
$⇒AC=2\sqrt[]{2}cm$
$⇒BC^2=AB^2+AC^2$
$⇒BC=\sqrt[]{(2\sqrt[]{2})^2+(2\sqrt[]{2})^2}=4cm$
$⇒CH=BC-BH=4-2=2cm$