Đáp án:
a)
Xét $\triangle ABC$ vuông tại $A$ có
$BC^2=AB^2+AC^2$ (đly pitago)
$\Rightarrow BC^2=18^2+24^2$
$\Rightarrow BC=30$
Ta có:
$\dfrac{AD}{BD}=\dfrac{AC}{BC}=\dfrac{24}{30}$
$\Rightarrow \dfrac{AD}{24}=\dfrac{BD}{30}=\dfrac{AD+BD}{24+30}=\dfrac{AB}{54}=\dfrac{18}{54}=\dfrac{1}{3}$
$\Rightarrow AD=8$
$\Rightarrow BD=10$
b)
Xét $\triangle HBA$ và $\triangle ABC$ có
$\widehat{AHB}=\widehat{BAC}=90^0$
$\hat{B}$ chung
$\Rightarrow \triangle HBA~\triangle ABC$ (g.g)
c)
Ta có:
$\widehat{KBD}+\widehat{KDB}=90^0$
$\widehat{ADC}+\widehat{ACD}=90^0$
mà $\widehat{KDB}=\widehat{ADC}$ (đối đỉnh)
$\Rightarrow \widehat{KBD}=\widehat{ACD}$
mà $\widehat{ACD}=\widehat{DCB}$ (do $CD$ là phân giác)
$\Rightarrow \widehat{KBD}=\widehat{DCB}$
Xét $\triangle KBD$ và $\triangle HCI$ có
$\widehat{BKD}=\widehat{IHC}=90^0$
$\widehat{KBD}=\widehat{DCB}$
$\Rightarrow \triangle KBD~\triangle HCI$ (g.g)
$\Rightarrow \dfrac{KB}{HC}=\dfrac{KD}{HI}$
$\Rightarrow KD.HC=KB.HI$
d)
Xét $\triangle BHE$ và $\triangle BKC$ có
$\widehat{B}$ chung
$\widehat{BHE}=\widehat{BKC}=90^0$
$\Rightarrow \triangle BHE~\triangle BKC$ (g.g)
$\Rightarrow \dfrac{BH}{BK}=\dfrac{BE}{BF}$ (1)
Lại có $\triangle HBA~\triangle ABC$ (cmt)
$\Rightarrow \dfrac{BH}{AB}=\dfrac{AB}{BC}$
$\Rightarrow AB^2=BH.BC$ (2)
Từ (1) và (2) suy ra $AB^2=BK.BE$
mà $AB=BF$ (gt)
$\Rightarrow BF^2=BK.BE$
$\Rightarrow \dfrac{BF}{BK}=\dfrac{BE}{BF}$
Xét $\triangle BFE$ và $\triangle BKF$ có
$\widehat{B}$ chung
$\dfrac{BF}{BK}=\dfrac{BE}{BF}$
$\Rightarrow \triangle BFE~\triangle BKF$ (c.g.c)
$\Rightarrow \widehat{BFE}=\widehat{BKF}=90^0$
$\Rightarrow BF\bot EF$
