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a) Xét 2 tam giác vuông \(\Delta EBC\)và \(\Delta DCB\)có:
\(BC:\)cạnh chung
\(\widehat{EBC}=\widehat{DCB}\)
suy ra: \(\Delta EBC=\Delta DCB\) (ch_gn)
\(\Rightarrow\)\(BD=EC\) (cạnh tương ứng)
b) \(\Delta ABC\)có các đường cao \(BD,EC\)cắt nhau tại \(H\)
\(\Rightarrow\)\(H\)là trực tâm của \(\Delta ABC\)
\(\Rightarrow\)\(AH\)là đường cao của \(\Delta ABC\)
\(\Rightarrow\)\(AH\perp BC\)
c) \(\Delta ABC\)cân tại A có AH là đường cao
nên AH đồng thời là đường phân giác
\(\Rightarrow\)\(\widehat{EAH}=\widehat{DAH}\) (đpcm)
a: Xét ΔBEC vuông tại E và ΔCDB vuông tại D có
BC chung
\(\widehat{EBC}=\widehat{DCB}\)
Do đó:ΔBEC=ΔCDB
b: Xét ΔABD vuông tại D và ΔACE vuông tại E có
AB=AC
\(\widehat{BAD}\) chung
Do đó:ΔABD=ΔACE
Suy ra: AD=AE
c: Ta có: ΔBEC=ΔCDB
nên \(\widehat{IBC}=\widehat{ICB}\)
hayΔIBC cân tại I
Xét ΔABI và ΔACI có
AB=AC
AI chung
BI=CI
Do đó:ΔABI=ΔACI
Suy ra: \(\widehat{BAI}=\widehat{CAI}\)
hay AI là tia phân giác của góc BAC
d: Xét ΔABC có AE/AB=AD/AC
nên DE//BC
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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
a, tg ADB và tg AEC có
^E1 = ^D1 = 90 độAB = AC
^A chung
=> tg ADB = tg AEC
=> AD = AE
=> tg ADE cân
b, tg ABI và tg ACI có
^E1 = ^D1 = 90 độ
AI chung
AB = AC
=> tg ABI = tg ACI
=> ^A1 = ^A2 ( góc t/ứ)
=> IB = IC ( cạnh t/ứ)
=> tg IBC cân
c, vì ^A1 = ^A2 ( câu b )
=> AI là tpg của góc EAD
a: Xét ΔABD vuông tại D và ΔACE vuông tại E có
AB=AC
\(\widehat{BAD}\) chung
Do đó: ΔABD=ΔACE
Suy ra; BD=CE
b: Xét ΔAEH vuông tại E và ΔADH vuông tại D có
AH chung
AE=AD
Do đó: ΔAEH=ΔADH
Suy ra: \(\widehat{EAH}=\widehat{DAH}\)
hay AH là tia phân giác của góc BAC
c: Xét ΔABC cso AE/AB=AD/AC
nên DE//BC
