Giải thích các bước giải:
a.Xét $\Delta DBH,\Delta DAC$ có:
$\widehat{HDB}=\widehat{ADC}(=90^o)$
$\widehat{BHD}=90^o-\widehat{HBD}=90^o-\widehat{EBC}=\widehat{ECB}=\widehat{ACD}$
$\to\Delta DBH\sim\Delta DAC(g.g)$
$\to\dfrac{DB}{DA}=\dfrac{DH}{DC}$
$\to DB.DC=DH.DA$
2.Xét $\Delta AEB,\Delta AFC$ có:
Chung $\hat A$
$\widehat{AEB}=\widehat{AFC}(=90^o)$
$\to\Delta AEB\sim\Delta AFC(g.g)$
$\to \dfrac{AE}{AF}=\dfrac{AB}{AC}$
$\to AE.AC=AF.AB$
3.Ta có:
$\widehat{MAC}=\widehat{MBC}=\widehat{EBC}=90^o-\widehat{ECB}=90^o-\widehat{ACD}=\widehat{CAD}=\widehat{CAH}$
$\to AC$ là phân giác $\widehat{HAM}$
$\to AE$ là phân giác $\widehat{HAM}$
Mà $BE\perp AC\to AE\perp HM$
$\to \Delta AHM$ cân tại $A\to M,H$ đối xứng qua $AE$
$\to M,H$ đối xứng qua $AC$
4.Ta có $\widehat{BEC}=\widehat{BFC}=90^o$
$\to BCEF$ nội tiếp
$\to \widehat{BEF}=\widehat{FCB}=\widehat{NCB}=\widehat{BMN}$
$\to EF//MN$
5.Gọi $AG$ là đường kính của $(O)$
$\to \widehat{ACG}=90^o=\widehat{ADB}$
Mà $\widehat{AGC}=\widehat{ABC}=\widehat{ABD}$
$\to\Delta ACG\sim\Delta ADB(g.g)$
$\to\dfrac{AC}{AD}=\dfrac{AG}{AB}$
$\to AB.AC=AD.AG=2R.AD$
6.Vì $BCEF$ nội tiếp
$\to\widehat{AEF}=\widehat{ABC},\widehat{AFE}=\widehat{ACB}$
$\to\Delta AEF\sim\Delta ABC(g.g)$
$\to \dfrac{EF}{BC}=\dfrac{AE}{AB}=\cos\widehat{BAC}$
$\to EF=BC\cos\widehat{BAC}$
7.Xét $\Delta BDH,\Delta BEC$ có:
Chung $\hat B$
$\widehat{BDH}=\widehat{BEC}(=90^o)$
$\to\Delta BDH\sim\Delta BEC(g.g)$
$\to\dfrac{BD}{BE}=\dfrac{BH}{BC}$
$\to BH.BE=BD.BC$
Tương tự $CH.CF=CD.CB$
$\to BH.BE+CH.CF=BD.BC+CD.CB=BC^2$
8.Ta có $BCEF$ nội tiếp
$\to \widehat{EBF}=\widehat{ECF}$
$\to\widehat{ABM}=\widehat{ACN}$
$\to AM=AN$
$\to A$ nằm giữa cung $MN$
$\to AO\perp OM$
$\to AO\perp EF$ vì $EF//MN$
