a/ Xét \(\Delta BAH;\Delta CAH\) có :
\(\left\{{}\begin{matrix}AB=AC\\\widehat{B}=\widehat{C}\\\widehat{AHB}=\widehat{AHC}=90^0\end{matrix}\right.\)
\(\Leftrightarrow\Delta BAH=\Delta CAH\left(c-g-c\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}HB=HC\\\widehat{BAH=}\widehat{CAH}\end{matrix}\right.\)
b/ \(HB+HC=BC\) (H nằm giữa B và C)
Mà \(HB=HC\)
\(\Leftrightarrow HB=HC=\dfrac{BC}{2}=\dfrac{8}{2}=4cm\)
Xét \(\Delta ABH\)có \(\widehat{AHB}=90^0\)
\(\Leftrightarrow AB^2=AH^2+BH^2\)(định lý Py ta go)
\(\Leftrightarrow5^2=AH^2+4^2\)
\(\Leftrightarrow AH^2=5^2-4^2\)
\(\Leftrightarrow AH^2=9\)
\(\Leftrightarrow AH=3cm\)
c/ Xét \(\Delta BHD;\Delta CHE\) có :
\(\left\{{}\begin{matrix}\widehat{BHA}=\widehat{AHC}=90^0\\HB=HC\\\widehat{B}=\widehat{C}\end{matrix}\right.\)
\(\Leftrightarrow\Delta BHD=\Delta CHE\left(ch-gn\right)\)
\(\Leftrightarrow HD=HE\)
\(\Leftrightarrow\Delta ADE\) cân tại H
a/ Xét ΔBAH;ΔCAHΔBAH;ΔCAH có :
⎧⎪ ⎪⎨⎪ ⎪⎩AB=ACˆB=ˆCˆAHB=ˆAHC=900{AB=ACB^=C^AHB^=AHC^=900
⇔ΔBAH=ΔCAH(c−g−c)⇔ΔBAH=ΔCAH(c−g−c)
⇔{HB=HCˆBAH=ˆCAH⇔{HB=HCBAH=^CAH^
b/ HB+HC=BCHB+HC=BC (H nằm giữa B và C)
Mà HB=HCHB=HC
⇔HB=HC=BC2=82=4cm⇔HB=HC=BC2=82=4cm
Xét ΔABHΔABHcó ˆAHB=900AHB^=900
⇔AB2=AH2+BH2⇔AB2=AH2+BH2(định lý Py ta go)
⇔52=AH2+42⇔52=AH2+42
⇔AH2=52−42⇔AH2=52−42
⇔AH2=9⇔AH2=9
⇔AH=3cm⇔AH=3cm
c/ Xét ΔBHD;ΔCHEΔBHD;ΔCHE có :
⎧⎪ ⎪⎨⎪ ⎪⎩ˆBHA=ˆAHC=900HB=HCˆB=ˆC{BHA^=AHC^=900HB=HCB^=C^
⇔ΔBHD=ΔCHE(ch−gn)⇔ΔBHD=ΔCHE(ch−gn)
⇔HD=HE⇔HD=HE
⇔ΔADE⇔ΔADE cân tại H