a)
Xét $\Delta AHI$ và $\Delta ABH$, ta có:
$\widehat{BAH}$ là góc chung
$\widehat{AIH}=\widehat{AHB}=90{}^\circ $
$\to \Delta AHI\backsim\Delta ABH\,\,\,\left( g.g \right)$
$\to \dfrac{AH}{AB}=\dfrac{AI}{AH}\,\,\,\to \,\,\,A{{H}^{2}}=AI.AB\,\,\,\left( 1 \right)$
Xét $\Delta AHK$ và $\Delta ACH$, ta có:
$\widehat{CAH}$ là góc chung
$\widehat{AKH}=\widehat{AHC}=90{}^\circ $
$\to \Delta AHK\backsim\Delta ACH\,\,\,\left( g.g \right)$
$\to \dfrac{AH}{AC}=\dfrac{AK}{AH}\,\,\,\to \,\,\,A{{H}^{2}}=AK.AC\,\,\,\left( 2 \right)$
Từ $\left( 1 \right)$ và $\left( 2 \right)$
$\Rightarrow AK.AC=AI.AB$
$\Rightarrow \dfrac{AK}{AB}=\dfrac{AI}{AC}$
Xét $\Delta AKI$ và $\Delta ABC$, ta có:
$\widehat{BAC}$ là góc chung
$\dfrac{AK}{AB}=\dfrac{AI}{AC}\,\,\,\left( cmt \right)$
$\to \Delta AKI\backsim\Delta ABC\,\,\,\left( c.g.c \right)$
b)
Xét $\Delta AHB$ và $\Delta CHA$, ta có:
$\widehat{AHB}=\widehat{CHA}=90{}^\circ $
$\widehat{ABH}=\widehat{CAH}$ ( cùng phụ $\widehat{ACB}$ )
$\to \Delta AHB\backsim\Delta CHA\,\,\,\left( g.g \right)$
$\to \dfrac{AH}{CH}=\dfrac{BH}{AH}$
$\to A{{H}^{2}}=BH.CH$
$\to A{{H}^{2}}=4.9$
$\to A{{H}^{2}}=36$
$\to AH=6\,\,\,\left( cm \right)$
Diện tích $\Delta ABC$:
${{S}_{\Delta ABC}}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.6.13=39\,\,\,\left( c{{m}^{2}} \right)$
c)
$\Delta AHB$ vuông tại $H$
$\to A{{B}^{2}}=B{{H}^{2}}+A{{H}^{2}}$ ( định lý Pi-ta-go )
$\to A{{B}^{2}}={{4}^{2}}+{{6}^{2}}$
$\to A{{B}^{2}}=52$
$\to AB=2\sqrt{13}\,\,\,\left( cm \right)$
$\Delta AHC$ vuông tại $H$
$\to A{{C}^{2}}=C{{H}^{2}}+A{{H}^{2}}$ ( định lý Pi-ta-go )
$\to A{{C}^{2}}={{9}^{2}}+{{6}^{2}}$
$\to A{{C}^{2}}=117$
$\to AC=3\sqrt{13}\,\,\,\left( cm \right)$
Ta có: $A{{H}^{2}}=AI.AB\,\,\,\left( cmt \right)$
$\to AI=\dfrac{A{{H}^{2}}}{AB}=\dfrac{{{6}^{2}}}{2\sqrt{13}}=\dfrac{18\sqrt{13}}{13}\,\,\,\left( cm \right)$
Ta có: $A{{H}^{2}}=AK.AC\,\,\,\left( cmt \right)$
$\to AK=\dfrac{A{{H}^{2}}}{AC}=\dfrac{{{6}^{2}}}{3\sqrt{13}}=\dfrac{12\sqrt{13}}{13}\,\,\,\left( cm \right)$
Xét tứ giác $AKHI$, ta có:
$\begin{cases}\widehat{IAK}=90{}^\circ\\\widehat{AIH}=90{}^\circ\\\widehat{AKH}=90{}^\circ\end{cases}$
$\to AKHI$ là hình chữ nhật
$\to {{S}_{\Delta AKHI}}=AI.AK=\dfrac{18\sqrt{13}}{13}\,.\,\dfrac{12\sqrt{13}}{13}=\dfrac{216}{13}\,\,\,\left( c{{m}^{2}} \right)$
