Lời giải:
a)
Xét tam giác $HBA$ và $ABC$ có:
\(\widehat{AHB}=\widehat{CAB}=90^0\)
\(\widehat{B}\) chung
\(\Rightarrow \triangle HBA\sim \triangle ABC(g.g)\)
b)
Xét tam giác $ABH$ và $CAH$ có:
\(\widehat{AHB}=\widehat{CHA}=90^0\)
\(\widehat{ABH}=\widehat{CAH}(=90^0-\widehat{BAH})\)
\(\Rightarrow \triangle ABH\sim \triangle CAH(g.g)\)
\(\Rightarrow \frac{AB}{BH}=\frac{CA}{AH}\Leftrightarrow \frac{2AB}{BH}=\frac{CA}{\frac{AH}{2}}\Leftrightarrow \frac{BD}{BH}=\frac{AC}{AM}\)
Xét tam giác $BHD$ và $AMC$ có:
\(\widehat{DBH}=\widehat{CAM}(=90^0-\widehat{BAH})\)
\(\frac{BD}{BH}=\frac{AC}{AM}\) (cmt)
\(\Rightarrow \triangle BHD\sim \triangle AMC(c.g.c)\Rightarrow \frac{BD}{HD}=\frac{AC}{MC}\Rightarrow BD.MC=HD.AC\)
Ta có đpcm.
c) Gọi $K,L$ là giao điểm $MC-DH$ và $AC-DH$
Vì \(\triangle BHD\sim \triangle AMC\Rightarrow \widehat{D_1}=\widehat{C_1}\)
Mà \(\widehat{L_1}=\widehat{L_2}\) (đối đỉnh)
\(\Rightarrow \widehat{D_1}+\widehat{L_1}=\widehat{C_1}+\widehat{L_2}\)
\(\Rightarrow 180^0-(\widehat{D_1}+\widehat{L_1})=180^0-(\widehat{C_1}+\widehat{L_2})\)
\(\Rightarrow \widehat{DAL}=\widehat{LKC}\Rightarrow \widehat{LKC}=90^0\)
\(\Rightarrow DH\perp MC\) (đpcm)
