a.Ta có: AH⊥BC→ΔAHB,ΔAHCAH⊥BC→ΔAHB,ΔAHC vuông tại HH

→AB2=AH2+BH2→AH2=AB2−BH2→AB2=AH2+BH2→AH2=AB2−BH2

      AC2=AH2+CH2→AH2=AC2−CH2AC2=AH2+CH2→AH2=AC2−CH2

→AB2−BH2=AC2−CH2→AB2−BH2=AC2−CH2

→AB2+CH2=AC2+BH2→AB2+CH2=AC2+BH2

b.Ta có: ΔABFΔABF vuông tại AA

→ˆAFB<90o→AFB^<90o

→ˆBFC=180o−ˆAFB>90o→BFC^=180o−AFB^>90o

→ΔFBC→ΔFBC tù tại FF

→FB<BC,FC<BC→FB<BC,FC<BC

Kẻ FD⊥BCFD⊥BC tại DD

Nếu EE nằm giữa D,BD,B

→DE<DB→DE<DB

→FE<FB<BC→FE<FB<BC

Nếu EE nằm giữa D,CD,C

→DE<DC→DE<DC

→FE<FC<BC→FE<FC<BC

c.Ta có: ΔABCΔABC vuông tại AA

→BC2=AB2+AC2=100→BC2=AB2+AC2=100

→BC=10→BC=10

Ta có: SABC=12AB⋅AC=24SABC=12AB⋅AC=24

Do AH⊥BCAH⊥BC

→12AH⋅BC=SABC=24→12AH⋅BC=SABC=24

→12AH⋅10=24→12AH⋅10=24

→AH=4.8→AH=4.8

→BH2=AB2−AH2=12.96→BH2=AB2−AH2=12.96

→BH=3.6→BH=3.6

Lại có: AC2=AH2+HC2AC2=AH2+HC2

→HC2=AC2−AH2=40.96→HC2=AC2−AH2=40.96

→CH=6.4→CH=6.4