Ta có:
$AM = MB =\dfrac12AB$
$AN = NH =\dfrac12AH$
$\to MN$ là đường trung bình của $∆ABH$
$\to MN =\dfrac12BH;\, MN//BH$
$\to MN =\dfrac12BH;\, MN\perp BH$
Ta được:
$S_{AHM}=\dfrac12MN.AH$
$\to S_{AHM}=\dfrac12\cdot\dfrac12BH.AH$
$\to S_{AHM}=\dfrac12S_{ABH}$
Tương tự, ta có:
$HB = HC =\dfrac12BC$
$\to HB.AH = \dfrac12BC.AH$
$\to \dfrac12HB.AH =\dfrac12\cdot\dfrac12BC.AH$
$\to S_{AHB}=\dfrac12S_{ABC}$