Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta AHC$ có:

$\widehat{AHB}=\widehat{AHC}(=90^o)$

$\widehat{HAB}=90^o-\widehat{HAC}=\widehat{HCA}$

$\to \Delta HAB\sim\Delta HCA(g.g)$

$\to \dfrac{HA}{HC}=\dfrac{HB}{AH}$

$\to CH=\dfrac{AH^2}{HB}=9$

$\to AB=\sqrt{AH^2+HB^2}=2\sqrt{13}, AC=\sqrt{AH^2+HC^2}=3\sqrt{13}$

b.Xét $\Delta AIH,\Delta AHB$ có:

Chung $\hat A$

$\widehat{AIH}=\widehat{AHB}(=90^o)$

$\to \Delta AIH\sim\Delta AHB(g.g)$

$\to \dfrac{AI}{AH}=\dfrac{AH}{AB}$

$\to AI\cdot AB=AH^2$

$\to AI\cdot AB+HC^2=AH^2+HC^2=AC^2$

c.Vì $HI\perp AB, HK\perp AC, AB\perp AC\to AIHK$ là hình chữ nhật $\to AH=IK$

Ta có: $S_{ABC}=2.5dm^2$

$\to \dfrac12AH\cdot BC=2.5\to AH\cdot BC=5$

Ta có: $S_{ACH}=1.6$

$\to \dfrac12AH\cdot HC=1.6$

$\to AH\cdot HC=3.2$

$to AH\cdot BC-AH\cdot HC=1.8$

$\to AH(BC-HC)=1.8$

$\to AH\cdot BH=1.8$

$\to AH\cdot BH\cdot AH\cdot CH=5.76$

$\to AH^2\cdot BH\cdot CH=5.76$

$\to AH^2\cdot AH^2=5.76$ (từ câu a $\to AH^2=HB\cdot HC$)

$\to AH^4=5.76$

$\to AH=\dfrac{2\sqrt{15}}5$

$\to IK=\dfrac{2\sqrt{15}}5$