From Ceva's Theorem, it is evident that $|AK|\cdot|BL|\cdot|CH|=|AL|\cdot|BH|\cdot|CK|$, since the altitudes are concurrent. Now, we can use trigonometry to show that $|AK|\cdot|BL|\cdot|CH|=|HK|\cdot|KL|\cdot|LH|$. Let $O$ be the orthocenter of $\triangle ABC$. We know that $\angle ACL=90^\circ-A$, so $\angle COK=A$. Now, using cyclic quadrilateral $OKCH$, we can find that $\angle KHC=A$ as well, which implies that $\angle AHK=90^\circ-A$. Now, by the Law of Sines on $\triangle AHK$, we know that $$\frac{HK}{\sin(90^\circ -C)}=\frac{AK}{\sin(90^\circ-A)}\implies HK=\frac{AK\cos C}{\cos A}.$$Due to symmetry, we know that $LH=\frac{CH\cos B}{\cos C}$ and $KL=\frac{BL\cos A}{\cos B}$. Thus, $$|HK|\cdot|KL|\cdot|LH|=\frac{|AK|\cdot|BL|\cdot|CH|\cos A\cos B\cos C}{\cos A\cos B\cos C}=|AK|\cdot|BL|\cdot|CH|.$$