Giải thích các bước giải:
1.Xét $\Delta OPB, \Delta AOC$ có:
$\widehat{BOP}=\widehat{AOC}$
$\widehat{OPB}=\widehat{ACO}$
$\to \Delta OPB \sim\Delta OAC(g.g)$
2.Xét $\Delta AKE, \Delta ACP$ có:
Chung $\hat A$
$\widehat{AEK}=\widehat{AED}=180^o-\widehat{DEC}=180^o-\widehat{DBC}=\widehat{ABC}=\widehat{APC}$
$\to \Delta AKE\sim\Delta ACP(g.g)$
$\to \dfrac{AK}{AC}=\dfrac{AE}{AP}$
$\to AK.AP=AC.AE$
3.Kẻ $AH$ là tiếp tuyến của $(O)\to OF\perp AF$
$\to AH^2=AO^2-OF^2=3R^2$
Xét $\Delta AEH,\Delta AHC$ có:
Chung $\hat A$
$\widehat{AHE}=\widehat{ACH}$
$\to \Delta AHE\sim\Delta ACH(g.g)$
$\to \dfrac{AF}{AC}=\dfrac{AE}{AH}$
$\to AE.AC=AH^2=3R^2$
$\to AK.AP=3R^2$
Từ câu a $\to \dfrac{OB}{OA}=\dfrac{OP}{OC}\to OP.OA=OB.OC=R^2$
$\to OP.2R=R^2\to OP=\dfrac12R$
$\to AP=AO+OP=\dfrac52R$
$\to AK.\dfrac52R=3R^2$
$\to AK=\dfrac65R^2$
$\to K$ cố định
$\to DE$ đi qua $K$ cố định
4.Gọi $(ADE)\cap AO=F$
Xét $\Delta AEF,\Delta AOC$ có:
Chung $\hat A$
$\widehat{AFE}=\widehat{ADE}=\widehat{BDE}=\widehat{BCE}=\widehat{ACO}$
$\to \Delta AEF\sim\Delta AOC(g.g)$
$\to \dfrac{AE}{AO}=\dfrac{AF}{AC}$
$\to AF.AO=AE.AC=3R^2$
$\to AF=\dfrac32R$
$\to F$ cố định
Ta có $O$ là trung điểm $BC$
$\to S_{ABC}=2S_{AOB}, S_{PBC}=2S_{BOP}$
$\to S_{ABPC}=2S_{ABP}$
Kẻ $BI\perp AO\to BI\le BO=R$
$\to S_{ABP}=\dfrac12BI\cdot AP\le \dfrac12R\cdot \dfrac52R=\dfrac54R^2$
$\to S_{ABPC}\le \dfrac52R^2$
Dấu = xảy ra khi $I\equiv O\to AO\perp BC$
