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Xét trường hợp đường thẳng d cắt cạnh BC tại 1 điểm . Tìm mối liên hệ giữa 3 đoạn thẳng BC , EC , DE
chi oi,kho qua em ko bit lam,k em nhe,hom nay em k chi qua 3 lan rui do
a, ta có: tam giác abc vuông cân tại a => góc abc=góc acb (1)
(.) bd ⊥d và ce ⊥d => bd // ce =>góc dbc=góc ecb (2)
(.)góc dba + abc =góc eca + acb (3)
từ (1),(2) và (3) => góc dba = eca
xét tam giác abd và tam giác cea có
ab=ac (gt) và góc d =góc e=90 độ (gt) và góc dba =eca(cmt)
=> tam giác abd = tam giác cea(cạnh huyền- góc nhọn)
mình hơi gấp có gì bạn tự kí hiệu nha
.Do ΔDBA=ΔAEC =>góc BAD=CAE
mặt khác :góc BAD+BAC+CAE=180 độ
mà góc BAC=90=>BAD+CAE=90
suy ra: góc BAD=CAE=90/2=45 độ
xét ΔAEC có: ^E=90,^EAC=45 =>^ACE=45 suy ra:ΔEAC vuông cân tại E=>AE=EC (1)
xét ΔDBA có: ^D=90, ^DAB=45 => ^DBA=45 suy ra:ΔDBA vuông cân tại D =>DB=DA (2)
.DA+AE=DE (3)
từ (1),(2) và (3) =>BD+CE=DE
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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