Dựng $AH\bot BC$
$\Rightarrow \left\{ \begin{array}{l} 16^2-AH^2=BH^2 \\ 20^2-AH^2=CH^2 \end{array} \right .$
Trừ vế với vế:
$20^2-16^2=HC^2-HB^2=(HC-HB)(HC+HB)=24(HC-HB)$
$\Rightarrow HC-HB=6$ và $HC+HB=24$
$\Rightarrow HB=9$, $HC=15$
$\Rightarrow AH=\sqrt{16^2-9^2}=5\sqrt7$
Dựng $IM\bot BC$
$\tan \widehat B=\dfrac{AH}{BH}=\dfrac{5\sqrt7}{9}$
$\widehat B=55,77\Rightarrow \widehat {\dfrac{B}{2}}=27,88$
$\Rightarrow \tan \widehat {\dfrac{B}{2}}=0,529=\dfrac{IM}{MB}$
Tương tự $\tan \widehat {\dfrac{C}{2}}=0,378=\dfrac{IM}{MC}$
Chia vế với vế: $\dfrac{MC}{MB}=\dfrac{0,529}{0,378}=1.4$
$\Rightarrow MC-1,4BM=0$ và $MC+MB=24$
$\Rightarrow MC=14$ và $MB=10$
$\Rightarrow IM=0,378.14=5,292$
$\Rightarrow KH=2IM=10,502$
Do $DE\parallel BC$
$\Rightarrow\dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{DE}{BC}=\dfrac{AK}{AH}=\dfrac{5\sqrt7-10,584}{5\sqrt7}=0.2 $
$\Rightarrow AD=0,2.AB=3,2$
$AE=0,2.AC=4$
$DE=4,8$
$P_{\Delta ADE}=AD+AE+DE=12$
