Đáp án: $AC=\sqrt{39},AB=\dfrac{2\sqrt{39}}{3}$
Giải thích các bước giải:
Ta có $S_{AHB}=4, S_{AHC}=9$
$\to\begin{cases}\dfrac12AH\cdot BH=4\\ \dfrac12AH\cdot HC=9\end{cases}$
$\to\begin{cases}AH\cdot BH=8\\ AH\cdot HC=18\end{cases}$
$\to AH\cdot BH\cdot AH\cdot HC=8\cdot 18$
$\to AH^2\cdot (BH\cdot HC)=144(*)$
Ta có $\widehat{AHB}=\widehat{AHC}=90^o,\widehat{BAH}=90^o-\widehat{HAC}=\widehat{HCA}$
$to \Delta AHB\sim\Delta CHA(g.g)$
$\to\dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to HA^2=HB\cdot HC$
Từ $(*)\to HA^4=144\to AH=2\sqrt{3}$
$\to HB\cdot HC=12$
Lại có $\dfrac{HB}{HC}=\dfrac{S_{AHB}}{S_{AHC}}=\dfrac49$
$\to HB=\dfrac49HC$
$\to HB\cdot HC=\dfrac49HC^2=12$
$\to HC=3\sqrt{3}$
$\to HB=\dfrac{4\sqrt{3}}{3}$
$\to AB=\sqrt{AH^2+HB^2}=\sqrt{(2\sqrt{3})^2+(\dfrac{4\sqrt{3}}{3})^2}=\dfrac{2\sqrt{39}}{3}$
$AC=\sqrt{AH^2+HC^2}=\sqrt{(2\sqrt{3})^2+(3\sqrt{3})^2}=\sqrt{39}$
