Giải thích các bước giải:

Bài 1:

a.Xét $\Delta EHI,\Delta FKI$ có:

$\hat H=\hat K(=90^o)$

$IE=IF$

$\widehat{EIH}=\widehat{FIK}$

$\to \Delta EIH=\Delta FIK$(cạnh huyền-góc nhọn)

$\to IH=IK$

b.Xét $\Delta IHF,\Delta IKE$ có:

$IH=IK$

$\widehat{HIF}=\widehat{EIK}$

$IF=IE$

$\to \Delta IHF=\Delta IKE(c.g.c)$

$\to \widehat{IHF}=\widehat{IKE}\to HF//KE$

     $HF=KE$

c.Ta có:

$EH\perp DI, FK\perp DI$

$\to DE>DH, DF>DK$

$\to DE+DF>DH+DK$

$\to DE+DF>(DI-IH)+(DI+IK)$

$\to DE+DF>2DI+(IK-IH)$

$\to DE+DF>2DI+0$

$\to DE+DF>2DI$

Bài 2:

a.Xét $\Delta ABM,\Delta CDM$có:

$MA=DM$

$\widehat{AMB}=\widehat{DMC}$

$MB=CM$

$\to \Delta AMB=\Delta DMC(c.g.c)$

b.Từ a $\to \widehat{MAC}=\widehat{MDB}$

$\to AC//BD$

c.Từ a $\to AC=BD,\widehat{MAC}=\widehat{MDB}$

Vì $AC>AB$

$\to BD>AB$

$\to \widehat{BAD}>\widehat{BDA}$

$\to \widehat{MAB}>\widehat{MDB}$

$\to \widehat{MAB}>\widehat{MAC}$