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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)
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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
a, tg ADB và tg AEC có
AB = 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
(g là góc)
Xét tg ABC,có:
AB=AC
=>tg ABC cân tại A
=>gABC = gACB
a)Xét tg BEC và tg CDB ,có:
BC:chung
gBEC =gCDB =90*(vì EC vuông gAB,BD vuông gAC)
gEBC = gDCB(cmt)
=>tg BEC = tg CDB(ch-gn)
=>BD=EC
b)Theo phần a,ta có:tg BEC = tg CDB(ch-gn)
=>gDBC=gECB(2 góc tương ứng)
=>tg BIC cân tại I
=>BI=CI
mà EI+IC=EC và DI+BI=BD(vì I là gđ của BD và EC) và BD=EC(theo phần a)
=>EI = DI
c)Xét tg ABC ,có:
AB=AC(gt)
BI=CI(cmt)
BH=CH(vì H là trung điểm của BC)
=>Ba điểm A, I, H thẳng hàng
(g là góc)
Xét tg ABC,có:
AB=AC
=>tg ABC cân tại A
=>gABC = gACB
a)Xét tg BEC và tg CDB ,có:
BC:chung
gBEC =gCDB =90*(vì EC vuông gAB,BD vuông gAC)
gEBC = gDCB(cmt)
=>tg BEC = tg CDB(ch-gn)
=>BD=EC
b)Theo phần a,ta có:tg BEC = tg CDB(ch-gn)
=>gDBC=gECB(2 góc tương ứng)
=>tg BIC cân tại I
=>BI=CI
mà EI+IC=EC và DI+BI=BD(vì I là gđ của BD và EC) và BD=EC(theo phần a)
=>EI = DI
c)Xét tg ABC ,có:
AB=AC(gt)
BI=CI(cmt)
BH=CH(vì H là trung điểm của BC)
=>Ba điểm A, I, H thẳng hàng
a) Xét tam giác vuông ADB và tam giác vuông ACE có:
Góc A chung
AB = AC (gt)
\(\Rightarrow\Delta ABD=\Delta ACE\) (Cạnh huyền - góc nhọn)
b) Do \(\Delta ABD=\Delta ACE\Rightarrow AD=AE\)
Xét tam giác vuông AEH và tam giác vuông ADH có:
Cạnh AH chung
AE = AD (cmt)
\(\Rightarrow\Delta AEH=\Delta ADH\) (Cạnh huyền - cạnh góc vuông)
\(\Rightarrow HE=HD\)
c) Xét tam giác ABC có BD, CE là đường cao nên chúng đồng quy tại trực tâm. Vậy H là trực tâm giác giác.
Lại có AM cũng là đường cao nên AM đi qua H.
d) Xét các tam giác vuông EBC và EAC, áp dụng định lý Pi-ta-go ta có:
\(BC^2=EB^2+EA^2;AC^2=EA^2+EC^2\)
Tam giác ABC cân tại A nên AB = AC hay \(AB^2=AC^2\)
Vậy nên \(AB^2+AC^2+BC^2=2AC^2+BC^2=2\left(EA^2+EC^2\right)+EB^2+EC^2\)
\(=3EC^2+2EA^2+BC^2\).