Giải thích các bước giải:
a.Ta có $10^2=6^2+8^2$
$\to BC^2=AB^2+AC^2$
$\to\Delta ABC$ vuông tại $A$
b.Ta có $I$ là giao ba đường phân giác
$\to \widehat{AIC}=180^o-\widehat{IAC}-\widehat{ICD}$
$\to \widehat{AIC}=180^o-\dfrac12\hat A-\dfrac12\hat C$
$\to \widehat{AIC}=180^o-\dfrac12(\hat A+\hat C)$
$\to \widehat{AIC}=180^o-\dfrac12(180^o-\hat B)$
$\to \widehat{AIC}=180^o-(90^o-\dfrac12\hat B)$
$\to \widehat{AIC}=90^o+\dfrac12\hat B$
$\to \widehat{AIC}=\widehat{BAD}+\widehat{ABD}$
$\to \widehat{AIC}=\widehat{BDC}$
c.Trên $BC$ lấy $K$ sao cho $BA=BK\to BK=6\to KC=BC-KB=4$
Xét $\Delta ABD,\Delta KBD$ có:
Chung $BD$
$\widehat{ABD}=\widehat{KBD}$
$BA=BK$
$\to\Delta ABD=\Delta KBD(c.g.c)$
$\to DA=DK, \widehat{DKB}=\widehat{DAB}=90^o\to DK\perp KC$
$\to DC^2=DK^2+KC^2$
$\to DC^2-DK^2=KC^2$
$\to (DC-DK)(DC+DK)=4^2$
$\to (DC-DK)(DC+DA)=16$
$\to (DC-DK).AC=16$
$\to (DC-DK).8=16$
$\to DC-DK=2$
$\to DC=2+DK$
$\to DC+DK=2+2DK$
$\to AC=2+2DK$
$\to 8=2+2DK$
$\to 2DK=6$
$\to DK=3$
$\to DA=DK=3$
$\to DC=AC-DA=5$
$\to 5AD=3DC$
d.Xét $\Delta CDI,\Delta CMI$ có:
Chung $CI$
$\widehat{DCI}=\widehat{CIM}$ vì $CI$ là phân giác góc $C$
$CD=CM=5$ do $M$ là trung điểm $BC\to MC=\dfrac12BC=5=CD$
$\to\Delta CDI=\Delta CMI(c.g.c)$
$\to \widehat{IMC}=\widehat{IDC}$
$\to \widehat{BMI}=180^o-\widehat{IMC}=180^o-\widehat{IDC}=\widehat{ADB}$
Lại có $\widehat{ABD}=\widehat{MBI}$ do $BI$ là phân giác góc $B$
$\to \widehat{BIM}=180^o-\widehat{IBM}-\widehat{IMB}=180^o-\widehat{DBA}-\widehat{ADB}=\widehat{BAD}=90^o$
$\to \Delta BIM$ vuông tại $I$
