Giải thích các bước giải:
1.Xét $\Delta AID,\Delta AIH$ có:
Chung $AI$
$\widehat{AID}=\widehat{AIH}(=90^o)$
$ID=IH$
$\to\Delta AID=\Delta AIH(c.g.c)$
b.Từ câu a
$\to AD=AH,\widehat{DAI}=\widehat{IAH}\to\widehat{DAB}=\widehat{BAH}$
Xét $\Delta ADB,\Delta AHB$ có:
Chung $AB$
$\widehat{DAB}=\widehat{BAH}$
$AD=AH$
$\to\Delta ADB=\Delta AHB(c.g.c)$
$\to\widehat{ADB}=\widehat{AHB}=90^o$
$\to BD\perp AD$
c.Tương tự câu b chứng minh được
$\Delta AEC=\Delta AHC(g.c.g), CE\perp AE$
$\to \widehat{HAC}=\widehat{CAE}$
$\to \widehat{HAE}=2\widehat{HAC}$
Mặt khác từ câu $b\to\widehat{DAB}=\widehat{BAH}\to \widehat{DAH}=2\widehat{BAH}$
$\to \widehat{DAE}=\widehat{DAH}+\widehat{HAE}=2\widehat{BAH}+2\widehat{HAC}=2\widehat{BAC}=180^o$
$\to D,A,E$ thẳng hàng
Mà $BD\perp AD, CE\perp AE\to BD\perp DE, CE\perp DE$
$\to BD//CE$