Câu 1:
$\cos \widehat{BAC}=\dfrac{AB}{AC}$$\Rightarrow AC=\dfrac{AB}{\cos \widehat{BAC}}=\dfrac{18}{\cos 23{}^\circ }\approx 19,6m$
Câu 2:
a)
Có: $\dfrac{1}{A{{H}^{2}}}=\dfrac{1}{A{{B}^{2}}}+\dfrac{1}{A{{C}^{2}}}$
$\Rightarrow AH=\dfrac{AB.AC}{\sqrt{A{{B}^{2}}+A{{C}^{2}}}}=\dfrac{6.8}{\sqrt{{{6}^{2}}+{{8}^{2}}}}=4,8cm$
b)
Có: $\begin{cases}AB^2=BH.BC\\AC^2=CH.BC\end{cases}\Rightarrow\dfrac{AB^2}{AC^2}=\dfrac{BH}{CH}\Rightarrow\dfrac{AB^2}{BH}=\dfrac{AC^2}{CH}$
c)
Có: $\begin{cases}\widehat{CAD}=\widehat{CAH}+\widehat{HAD}\\\widehat{CDA}=\widehat{CBA}+\widehat{BAD}\end{cases}$
Mà: $\begin{cases}\widehat{CAH}=\widehat{CBA}\\\widehat{HAD}=\widehat{BAD}\end{cases}$
$\Rightarrow \widehat{CAD}=\widehat{CDA}\Rightarrow \Delta ACD$ cân tại $C$$\Rightarrow AC=DC$
Có $\Delta AHC\backsim\Delta BAC\left( g.g \right)\Rightarrow \dfrac{AH}{AB}=\dfrac{HC}{AC}=\dfrac{HC}{DC}$
Theo tính chất phân giác, có: $\dfrac{AH}{AB}=\dfrac{DH}{BD}$
$\Rightarrow \dfrac{DH}{BD}=\dfrac{HC}{DC}\Rightarrow DH.DC=BD.HC$
