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a: Xét ΔABD và ΔEBD có
BA=BE
\(\widehat{ABD}=\widehat{EBD}\)
BD chung
Do đó: ΔABD=ΔEBD
b: Ta có: ΔABD=ΔEBD
nên \(\widehat{BAD}=\widehat{BED}=90^0\)
hay DE\(\perp\)BC
c: Xét ΔADK vuông tại A và ΔEDC vuông tại E có
DA=DE
\(\widehat{ADK}=\widehat{EDC}\)
Do đó: ΔADK=ΔEDC
Suy ra: AK=EC
Ta có: BA+AK=BK
BE+EC=BC
mà BA=BE
và AK=EC
nên BK=BC
a) Áp dụng pytago .
b) Xét t/g ABE; tg DBE:
AB = DB ( gt)
g ABE = DBE (suy từ gt)
BE chung
=> tg ABE = tg DBE (c.g.c)
c) Vì tg ABE = tg DBE (câu b)
=> AE = DE
Xét tg AEF ⊥⊥ tại A; tg DEC ⊥⊥ tại D:
AE = DE (c/m trên)
g AEF = g DEC (đối đỉnh)
=> tg AEF = tg DEC (cgv - gn)
=> EF = EC
d) Do tg AEF = tg DEC (câu c)
=> AE = DE
=> E ∈∈ đg trung trực của AD (1)
Lại do AB = BD (gt)
=> B ∈ đg trung trực của AD (2)
Từ (1) và (2) => BE là đg trung trực của AD.
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a, Xét tg BAE và tg BDE ( \(\widehat{BAE}=\widehat{BDE}=90^0\))
BA=BD (gt)
BE chung
=> tg BAE = tg BDE ( ch-cgv)
=> AE=ED
Ta có \(\hept{\begin{cases}BA=BD\left(gt\right)\\AE=ED\left(cmt\right)\end{cases}}< =>\)BE trung trực AD (đpcm)
b, +ED vuông BC
+ AH vuông BC
=> AH//DE
=> \(\widehat{HAD}=\widehat{ADE}\)( So le trong) (2)
Lại có gọi m là giao 2 đường thẳng BE và AD
vì BE trung trực AD =>+ \(\widehat{AME}=\widehat{EMD}=90^{0^{ }}\)
Xét tg AEM và tg DEM có \(\left(\widehat{AME}=\widehat{EMD}=90^0\left(cmt\right)\right)\)
+ AD = ED (cma)
+ EM chung
=> tg AEM = tg DEM ( ch-cgv)
=> \(\widehat{DAE}=\widehat{ADE}\)(2)
tỪ (1) VÀ (2) => \(\widehat{HAD}=\widehat{DAE}\)=> AD phân giác góc AHC
a) Ta thấy \(\widehat{ECN}=\widehat{ACB}\) (Hai góc đối đỉnh)
Tam giác ABC cân tại A nên \(\widehat{ACB}=\widehat{ABC}\Rightarrow\widehat{ECN}=\widehat{DBM}\)
Xét tam giác vuông BDM và CEN có:
BD = CE
\(\widehat{ECN}=\widehat{DBM}\) (cmt)
\(\Rightarrow\Delta BDM=\Delta CEN\) (Cạnh góc vuông và góc nhọn kề)
\(\Rightarrow BM=CN\) (Hai cạnh tương ứng)
b) Do \(\Delta BDM=\Delta CEN\Rightarrow MD=NE\)
Ta thấy MD và NE cùng vuông góc BC nên MD // NE
Suy ra \(\widehat{DMI}=\widehat{ENI}\) (Hai góc so le trong)
Xét tam giác vuông MDI và NEI có:
MD = NE
\(\widehat{DMI}=\widehat{ENI}\)
\(\Rightarrow\Delta MDI=\Delta NEI\) (Cạnh góc vuông và góc nhọn kề)
\(\Rightarrow MI=NI\)
Xét tam giác KMN có KI là đường cao đồng thời trung tuyến nên KMN là tam giác cân tại K.
c) Ta có ngay \(\Delta ABK=\Delta ACK\left(c-g-c\right)\Rightarrow\widehat{ABK}=\widehat{ACK}\) (1) và BK = CK
Xét tam giác BMK và CNK có:
BM = CN (cma)
MK = NK (cmb)
BK = CK (cmt)
\(\Rightarrow\Delta BMK=\Delta CNK\left(c-g-c\right)\Rightarrow\widehat{MBK}=\widehat{NCK}\) (2)
Từ (1) và (2) suy ra \(\widehat{ACK}=\widehat{NCK}\)
Chúng lại là hai góc kề bù nên \(\widehat{ACK}=\widehat{NCK}=90^o\)
Vậy \(KC\perp AN\)
