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Câu 1:
d A B C D E
Vì BD \(\perp\) d nên \(\widehat{BDA}\) = 90o
Ta có:
\(\widehat{BAD}\) + \(\widehat{BAC}\) + \(\widehat{CAE}\) = 180o
=> \(\widehat{BAD}\) + 90o + \(\widehat{CAE}\) = 180o
=> \(\widehat{BAD}\) + \(\widehat{CAE}\) = 90o (1)
Áp dụng tính chất tam giác vuông ta có:
\(\widehat{DBA}\) + \(\widehat{BAD}\) = 90o (2)
Từ (1) và (2) suy ra:
\(\widehat{BAD}\) + \(\widehat{CAE}\) = \(\widehat{DBA}\) + \(\widehat{BAD}\)
=> \(\widehat{CAE}\) = \(\widehat{DBA}\)
Xét \(\Delta\)DBA vuông tại D và \(\Delta\)EAC vuông tại E có:
BA = AC (giả thiết)
\(\widehat{DBA}\) = \(\widehat{EAC}\) (chứng minh trên)
=> \(\Delta\)DBA = \(\Delta\)EAC (cạnh huyền - góc nhọn)
=> DB = EA và DA = EC (2 cặp cạnh tương ứng).
Câu 2: Mk sẽ làm ở đây: /hoidap/question/166568.html
A E D M B N C
a) Xét \(\Delta\)ABM và \(\Delta\)CDM có:
AM = CM (suy từ giả thiết)
\(\widehat{AMB}\) = \(\widehat{CMD}\) (đối đỉnh)
BM = DM (giả thiết)
=> \(\Delta\)ABM = \(\Delta\)CDM (c.g.c)
b) Xét \(\Delta\)AMD và \(\Delta\)CMB có:
AM = CM (suy từ gt)
\(\widehat{AMD}\) = \(\widehat{CMB}\) (đối đỉnh)
MD = MB (gt)
=> \(\Delta\)AMD = \(\Delta\)CMB (c.g.c)
=> \(\widehat{ADM}\) = \(\widehat{CBM}\) (2 góc tương ứng)
mà 2 góc ở vị trí so le trong nên AD // BC.
c) Vì \(\Delta\)AMD = \(\Delta\)CMB (câu b)
nên \(\widehat{ADM}\) = \(\widehat{CBM}\) (2 góc tương ứng)
hay \(\widehat{EDM}\) = \(\widehat{NBM}\)
Xét \(\Delta\)EDM và \(\Delta\)NBM có:
\(\widehat{EDM}\) = \(\widehat{NBM}\) (chứng minh trên)
DM = BM (gt)
\(\widehat{EMD}\) = \(\widehat{NMB}\) (đối đỉnh)
=> \(\Delta\)EDM = \(\Delta\)NBM (g.c.g)
=> EM = NM (2 cạnh tương ứng)
Do đó M là trung điểm của NE.
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CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCGCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
A B C D E
a, xét ΔABC và ΔADE có : AD = AB (gt)
AE = AC (gt)
^BAC = ^DAE = 90
=> ΔABC = ΔADE (2cgv)
=> DE = BC (định nghĩa)
b, xét ΔEAC có ^EAC = 90
AE = AC (gt)
=> ΔEAC vuông cân tại A (định nghĩa)
=> ^CEA = 45 (tính chất) (1)
xét ΔBAD có ^BAD = 90
AD = AB (gt)
=> ΔBAD vuông cân tại A (định nghĩa)
=> ^ABD = 45 (2)
(1)(2) => ^CEA = ^ABD mà 2 góc này so le trong
=> BD // CE (định lí)
A B C D E
Xét tam giác BAC và tam giác DAE
có AB=AD (GT)
góc BAC = góc DAE = 900
AC=AE (GT)
suy ra tam giác BAC = tam giác DAE ( c.g.c)
suy ra BC= DE (hai cạnh tương ứng)
b) Vì AD=AB nên tam giác ABC cân tại A
mà góc A=900
suy ra tam giác ABC vuông cân tại A suy ra góc ABD=góc ADB=450 (1)
Xét tam giác ACE có AC=AE, góc CAE=900
suy ra tam giác ACE cân tại A suy ra góc ACE=góc AEC=450 (2)
Từ( 1) và (2) suy ra góc ABD= góc AEC (3)
mà góc ABD đồng vị với góc AEC (4)
Từ (3) và (4) suy ra BD//CE