$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{1.\ Xét\ \Delta ABD\ và\ \Delta ACE\ có:}\\ \mathrm{AB=AE\ ( \ do\ \Delta \ ABC\ cân\ tại\ A.)}\\ \mathrm{\widehat{ABD} =\ \widehat{ACE}}\\ \mathrm{BD=CE\ ( gt)}\\ \mathrm{\Rightarrow \Delta ABD\ =\ \Delta ACE\ ( c.g.c)}\\ \mathrm{\Rightarrow AD=AE}\\ \mathrm{\Rightarrow \Delta ADE\ cân\ tại\ A}\\ \mathrm{2.\ Xét\ \Delta DHB\ và\ \Delta CKE}\\ \mathrm{\widehat{DHB} =\ \widehat{EKC} =90^{o}}\\ \mathrm{\widehat{HBD} =\ \widehat{KCE} \ ( =\widehat{ABC} =\ \widehat{ACB})}\\ \mathrm{DB=CE\ ( gt)}\\ \mathrm{\Rightarrow \Delta DHB\ =\ \Delta CKE}\\ \mathrm{\Rightarrow DH=EK\ và\ \widehat{BDH} =\widehat{CEK}}\\ \mathrm{3.\ Xét\ \Delta ADH\ và\ \Delta AEK\ có}\\ \mathrm{AD=AE\ ( cmt)}\\ \mathrm{AH=AK}\\ \mathrm{DH=EK( cmt)}\\ \mathrm{\Rightarrow \Delta DHA\ =\ \Delta AKE\ ( c.c.c)}\\ \mathrm{4.Xét\ \Delta AHK\ có\ AH=AK}\\ \mathrm{\Rightarrow \Delta AHK\ \ cân\ \ tại\ A\ mà\ \Delta ABC\ \ cân\ tại\ A}\\ \mathrm{\ \Rightarrow \widehat{BAC} =\frac{\mathrm{\widehat{AHK}}}{2} =\frac{\mathrm{\widehat{ABC}}}{2}}\\ \mathrm{\Rightarrow \widehat{AHK} =\widehat{ABC}}\\ \mathrm{\Rightarrow HK//CB\ hay\ HK//DE}\\ \mathrm{5.\ Xét\ \Delta DEO\ có\ \widehat{BDH} =\widehat{CEK} \ ( cmt)}\\ \mathrm{\Rightarrow \Delta DEO\ cân\ tại\ O}\\ \mathrm{6.\ Xét\ \Delta AHO\ và\ \Delta AKO\ có:}\\ \mathrm{AO-chung}\\ \mathrm{AH=AK\ ( cmt)}\\ \mathrm{OH=OK\ ( \Delta DEO\ cân\ tại\ O)}\\ \mathrm{\Rightarrow \Delta AHO\ =\ \Delta AKO\ }\\ \mathrm{\Rightarrow \ \widehat{HAO} =\ \widehat{KAO}}\\ \mathrm{mà\ \ \widehat{DAH} =\ \widehat{EAK} \ ( \Delta DHA\ =\ \Delta AKE\ )}\\ \mathrm{\Rightarrow \widehat{DAO\ } \ =\widehat{EAO} \ }\\ \mathrm{\Rightarrow AO\ là\ p/g\ góc\ DAE}\\ \mathrm{7.Xét\ \Delta MPB\ và\ \Delta NQC\ có:}\\ \mathrm{MB=NC=\frac{1}{2} AC}\\ \mathrm{\widehat{MBP} =\widehat{NCQ}}\\ \mathrm{BP=CQ\ ( gt)}\\ \mathrm{\Rightarrow \Delta MPB\ =\ \Delta NQC}\\ \mathrm{\Rightarrow MP=NQ}\\ \mathrm{8.Xét\ \Delta AMF\ và\ \Delta ANF\ có:}\\ \mathrm{AF-chung}\\ \mathrm{\widehat{MAF} =\widehat{NAF} \ ( \ do\ \Delta AHO\ =\ \Delta AKO\ )}\\ \mathrm{AM=AN=\frac{1}{2} AB}\\ \mathrm{\Rightarrow \Delta AMF\ =\ \Delta ANF}\\ \mathrm{\Rightarrow MF=NF}\\ \mathrm{Xét\ \Delta PMQ\ và\ \Delta PNQ\ có:}\\ \mathrm{PQ-chung}\\ \mathrm{\widehat{MPQ} =\widehat{NPQ} \ \left( =180^{o} -\ \widehat{MPB}\right)}\\ \mathrm{MP=NQ\ ( cmt)}\\ \mathrm{\Rightarrow \Delta PMQ\ và\ \Delta PNQ}\\ \mathrm{\Rightarrow \widehat{PMQ} =\widehat{PNQ}}\\ \mathrm{Xét\ \Delta MFP\ và\ \Delta NFQ\ có:}\\ \mathrm{MF=NF\ ( cmt)}\\ \mathrm{\widehat{PMQ} =\widehat{PNQ} \ ( cmt)}\\ \mathrm{MP=NQ\ ( cmt)}\\ \mathrm{\Rightarrow \Delta MFP\ =\ \Delta NFQ}\\ \mathrm{9.\Delta ABC\ cân\ tại\ A\ có\ AO\ là\ đường\ phân\ giác}\\ \mathrm{\Rightarrow AO\ cũng\ làđường\ trung\ tuyến\ \ }\\ \mathrm{\Rightarrow A,I,O\ thẳng\ hàng\ ( 1)}\\ \mathrm{Cm\ \Delta ABN\ =\ \Delta ACM}\\ \mathrm{Cm\ \Delta BGC\ cân\ tại\ G\Rightarrow A,G,I\ thẳng\ hàng\ ( 2)}\\ \mathrm{Cm\ \ \Delta PFQ\ cân\ tại\ F\Rightarrow FI\ là\ đường\ trung\ trực\ của\ PQ\ cũng\ như\ BC}\\ \mathrm{mà\ AI\ là\ đường\ trung\ trực\ của\ BC}\\ \mathrm{\Rightarrow A,F,I\ thẳng\ hàng\ ( 3)}\\ \mathrm{Từ\ ( 1) \ ( 2) \ ( 3) \Rightarrow A,G,F,I,O\ thẳng\ hàng}\\ \mathrm{10.}\\ \mathrm{Ta\ có:\ AG+AM >MG\ ( T.c\ cạnh\ của\ tam\ giác)}\\ \mathrm{\Rightarrow AB+2AG >2MG}\\ \mathrm{\Rightarrow AC+BC+2GA >2MG}\\ \mathrm{\Rightarrow 2( AC+BC) +4GA >4GM}\\ \mathrm{Dễ\ thấy\ 2( AC+BC) >BC}\\ \mathrm{\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4GA >GA}\\ \mathrm{\Rightarrow BC+AG >4GM} \end{array}$

Ta có: 

AG+AM>MG (T.c cạnh của tam giác)

=>AB+2AG>2MG

=>AC+BC+2GA>2MG

=>2(AC+BC)+4GA>4GM

=>2(AC+BC)>BC           

=>4GA>GA

⇒BC+AG>4GM