a.Ta có: ˆAHB=ˆBAC=90o,ˆABH=ˆABCAHB^=BAC^=90o,ABH^=ABC^

→ΔAHB∼ΔCAB(g.g)→ΔAHB∼ΔCAB(g.g)

→ABCB=HBAB→ABCB=HBAB

→AB2=BH.BC→AB2=BH.BC

b.Ta có:

ˆAHB=ˆAHC=90o,ˆBAH=90o−ˆHAC=ˆACHAHB^=AHC^=90o,BAH^=90o−HAC^=ACH^

→ΔAHB∼ΔCHA(g.g)→ΔAHB∼ΔCHA(g.g)

→AHCH=HBHA→AHCH=HBHA

→AH2=HB.HC→AH2=HB.HC

c.Ta có: ˆAEH=ˆAHB=90o,ˆEAH=ˆBAHAEH^=AHB^=90o,EAH^=BAH^

→ΔAEH∼ΔAHB(g.g)→ΔAEH∼ΔAHB(g.g)

→AEAH=AHAB→AEAH=AHAB

→AH2=AE.AB→AH2=AE.AB

Tương tự AH2=AF.ACAH2=AF.AC

→AE.AB=AF.AC→AE.AB=AF.AC

→AEAC=AFAB→AEAC=AFAB

Mà ˆEAF=ˆBACEAF^=BAC^

→ΔAFE∼ΔABC(c.g.c)→ΔAFE∼ΔABC(c.g.c)

c.Ta có: SABC=12⋅BC⋅AH=180SABC=12⋅BC⋅AH=180

Ta có HE⊥AB,HF⊥AC,AB⊥AC→AEHFHE⊥AB,HF⊥AC,AB⊥AC→AEHF là hình chữ nhật

→EF=AH=12→EF=AH=12

Mà ΔAFE∼ΔABCΔAFE∼ΔABC

→SAEFSABC=(EFBC)2=425→SAEFSABC=(EFBC)2=425

→SAEF=425SABC=1445