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a) Xét ∆BEA và ∆CDA, ta có:
BA = CA (gt)
ˆAA^chung
AE = AD (gt)
Suy ra: ∆BEA = ∆CDA (c.g.c)
Vậy BE = CD (hai cạnh tương ứng)
b) ∆BEA = ∆CDA (chứng minh trên)
⇒\(\widehat{\text{B1}}=\widehat{\text{C1}}\);ˆE1=ˆD1E1^=D1^ (hai góc tương ứng)
ˆE1+ˆE2E1^+E2^=180o (hai góc kề bù)
ˆD1+ˆD2D1^+D2^=180o (hai góc kề bù)
Suy ra: ˆE2=ˆD2E2^=D2^
AB = AC (gt)
Ta có: \(CD=\frac12\times CA\)
=>\(S_{BDC}=\frac12\times S_{ABC}=\frac12\times180=90\left(\operatorname{cm}^2\right)\)
Ta có: BF=FE=EC
mà BF+FE+EC=BC
nên \(BF=FE=EC=\frac{BC}{3}\)
=>\(S_{DFE}=\frac{S_{DBC}}{3}=\frac{90}{3}=30\left(\operatorname{cm}^2\right)\)
CPE = 1/3 CPB = 1/3 CPA=1/4 CAE=1/8 ABC
BND=1/2 BNA=1/6 BNC=1/7 BCD=1/14ABC
AMF=1/4 AMC=1/8 ABM= 1/9 ABF=1/36 ABC
AMND=ABF – BND – AMF
=1/4 ABC = 1/14 ABC = 1/36 ABC= 7/42 ABC
BEPD= BCD = CPE
= ½ ABC – 1/8 ABC = 3/8 ABC
MNP = ABC – AEC – BEPD – AMND
= ABC – 1/3 ABC – 3/8 ABC – 7/42 ABC
= 1/8 ABC
Ta có: CE+EB=CB
=>\(EB=BC-\frac23\times BC=\frac13\times BC\)
=>CE=2xEB
=>\(EB=\frac12\times EC\)
=>\(S_{AEB}=\frac12\times S_{AEC};S_{OEB}=\frac12\times S_{OEC}\)
=>\(S_{AEB}-S_{OEB}=\frac12\times\left(S_{AEC}-S_{OEC}\right)\)
=>\(S_{AOB}=\frac12\times S_{AOC}\) (1)
DA+DB=AB
=>\(DB=AB-\frac13\times AB=\frac23\times AB\)
=>\(DB=2\times DA\)
=>\(S_{CDB}=2\times S_{CDA};S_{ODB}=2\times S_{ODA}\)
=>\(S_{CDB}-S_{ODB}=2\times\left(S_{CDA}-S_{ODA}\right)\)
=>\(S_{COB}=2\times S_{AOC}\) (2)
Từ (1),(2) suy ra \(\frac{S_{BOA}}{S_{BOC}}=\frac12:2=\frac14\)
=>\(S_{BOA}=\frac14\times S_{BOC}\)
Vì \(EB=\frac13\times BC\)
nên \(S_{OEB}=\frac13\times S_{BOC}\)
=>\(\frac{S_{EBO}}{S_{BOA}}=\frac13:\frac14=\frac43\)
=>\(\frac{OE}{OA}=\frac43\)
=>\(OE=\frac43\times OA\)
Ta có: OA+OE=AE
=>\(AE=OA+\frac43\times OA=\frac73\times OA\)
=>\(AO=\frac37\times AE\)
=>\(S_{ADO}=\frac37\times S_{ADE}\)
Vì \(AD=\frac13\times AB\) nên \(S_{ADE}=\frac13\times S_{AEB}\)
=>\(S_{ADO}=\frac37\times\frac13\times S_{AEB}=\frac17\times S_{AEB}\)
Vì BE=1/3BC
nên \(S_{AEB}=\frac13\times S_{ABC}\)
=>\(S_{ADO}=\frac17\times\frac13\times S_{ABC}=\frac{1}{21}\times S_{ABC}\)
Vì \(CE=\frac23\times CB\)
nên \(S_{CEO}=\frac23\times S_{COB}\)
Ta có: \(S_{AOB}+S_{AOC}+S_{BOC}=S_{ABC}\)
=>\(S_{ABC}=S_{AOC}+2\times S_{AOC}+\frac12\times S_{AOC}=\frac72\times S_{AOC}\)
=>\(\frac{S_{COB}}{S_{ABC}}=2:\frac72=\frac47\)
=>\(S_{BOC}=\frac47\times S_{ABC}\)
=>\(S_{CEO}=\frac23\times\frac47\times S_{ABC}=\frac{8}{21}\times S_{ABC}\)
=>\(\frac{S_{ADO}}{S_{CEO}}=\frac{1}{21}:\frac{8}{21}=\frac18\)