$\Delta ABC$ vuông tại $A$ có đường cao $AH$
$=>AH^2=BH.CH;\\ S_{\Delta ABH}=\dfrac{1}{2}AH.BH\\ S_{\Delta ACH}=\dfrac{1}{2}AH.CH\\ S_{\Delta ABH}.S_{\Delta ACH}=\dfrac{1}{4}AH.BH.AH.CH=\dfrac{1}{4}AH^4=19,44.34,56=>AH=7,2\\ =>BH=5,4;CH=9,6\\ =>AB=\sqrt{AH^2+BH^2}=9;AC=\sqrt{AH^2+BH^2}=12$