Giải thích các bước giải:
a.Xét $\Delta ADI,\Delta AHI$ 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 $AB$
$\widehat{DAB}=\widehat{BAH}$
$AD=AH$
$\to \Delta ABD=\Delta ABH(c.g.c)$
$\to \widehat{ADB}=\widehat{AHB}=90^o\to AD\perp rBD$
c.Ta có $BC=BH+HC=25$ vì $BH=9, HC=16$
Vì $\Delta ABC$ vuông tại $A$
$\to BC^2=AB^2+AC^2$
$\to BC^2=(AH^2+HB^2)+(AH^2+HC^2)=2AH^2+HB^2+HC^2$
$\to 2AH^2=BC^2-HB^2-HC^2=288$
$\to AH^2=144$
$\to AH=12$
d.Tương tự câu b chứng minh được $\Delta AHC=\Delta AEC$
$\to AH=AE, CH=CE, \widehat{HAC}=\widehat{CAE}$
$\to \widehat{DAE}=\widehat{DAH}+\widehat{HAE}=(\widehat{DAB}+\widehat{BAH})+(\widehat{HAC}+\widehat{CAE})$
$\to \widehat{DAE}=2\widehat{BAH}+2\widehat{HAC}$
$\to \widehat{DAE}=2(\widehat{BAH}+\widehat{HAC})$
$\to \widehat{DAE}=2\widehat{BAC}$
$\to \widehat{DAE}=180^o$
$\to A, D, E$ thẳng hàng
Mặt khác $\widehat{CEA}=\widehat{AHC}=90^o$
$\to CE\perp DE$
Kẻ $BF\perp CE=F$
Ta có $BD\perp AD\to BD\perp DE\to BD//EF$
Mà $DE//BF(\perp CE)$
$\to DE=BF, DB=EF$ (tính chất đoạn chắn)
Lại có $BF\perp CE\to BF\le BC=BH+HC=BD+CE$
$\to DE\le BD+CE$
