Ta có : AH⊥BC→ˆBHA=ˆBACAH⊥BC→BHA^=BAC^ 

→ΔABC∼ΔHBA(g.g)→ΔABC∼ΔHBA(g.g)

b.Từ câu a→ABHB=BCAB→AB2=BH.BC→AB=2√13→ABHB=BCAB→AB2=BH.BC→AB=213

Lại có AH⊥BC→AH=√AB2−BH2=6AH⊥BC→AH=AB2−BH2=6

c.Ta có : HE⊥HF→ˆFHC=ˆEHA(+ˆAHF=90o)HE⊥HF→FHC^=EHA^(+AHF^=90o)

Mà ˆEAH=ˆFCH(câub)→ΔHAE∼ΔHCF(g.g)EAH^=FCH^(câub)→ΔHAE∼ΔHCF(g.g)

→AECF=AHCH→AE.CH=AH.FC→AECF=AHCH→AE.CH=AH.FC

d.Từ câu c →AECF=AHCH→ΔHEF∼ΔHAC(c.g.c)→AECF=AHCH→ΔHEF∼ΔHAC(c.g.c)

→EFAC=HEHA≥HDHA(HD⊥AB=D)→EFAC=HEHA≥HDHA(HD⊥AB=D)

→EF≥AC.HDHA→EF≥AC.HDHA

Dấu = xảy ra khi E≡D→HE⊥ABE≡D→HE⊥AB