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Cho tam giác ABC vuông tại A. Gọi D là trung điểm của AB. Kẻ DE vuông góc với BC( E thuộc BC ) .Tính độ dài AC biết BE=7cm, EC=25cm
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Cho tam giác ABC vuông tại A, gọi D là trung điểm của AB. kẻ DE vuông góc với BC (E thuộc BC). Tính AC biết: BE = 7cm, EC = 25cm.
Cho tam giác ABC vuông tại A. Gọi D là trung điểm của AB. Kẻ DE vuông góc với BC( E thuộc BC ) .Tính độ dài AC biết BE=7cm, EC=25cm
Giúp mk vs nha các bn. Mk cần gấp!!!
Cho tam giác ABC vuông tại A. D là trung điểm của AB, kẻ DE vuông góc với BC (E thuộc BC) .Tính độ dài AC biết BE = 7cm, AC = 25cm
Cho tam giác ABC vuông tại A. Gọi D là trung điểm của AB. Kẻ DE vuông góc với BC (E thuộc BC). Tính độ dài AC, biết BE = 7 cm, EC = 25 cm
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cho tam giác abc vuông tại a gọi d là trung điểm ab kẻ de vuông góc với bc (e thuộc bc) tính độ dài ac biết be =7 ec = 25 cm
cho tam giác ABC vuông tại A. Gọi D alf truong điểm của AB. Kẻ DE vuông góc với BC (E\(\in\)BC). Tính AC, biết BE=7 cm, EC= 25cm
Ai giải giúp mình bài này với minh cần gấp; Cho tam giác ABC vuông tại A, đường phân giác BE (E thuộc AC) . Kẻ EH vuông góc BC (H thuộc BC). Gọi K là giao điểm của AB và HE. biết AB=6cm và BC=10cm. Câu a: Tính độ dài AC . Câu b: Chứng minh AB=HB, AE<EC . Câu c: chứng minh BE vuông góc CK và AH song song KC. Câu d: nếu góc ABC= 60 độ thì tam giác BHA là tam giác gì vì sao
Cho tam giác ABC vuông tại A,đườg phân giác BE (E thuộc AC) kẻ EH vuông góc với BC (H thuộc BC). Gọi K là giao điểm của AB và HE
a] Tính độ dài AC,biết AB= 6cm, BC = 10cm
b] Chứng minh AH=HB, AE <EC
c] Chứng minh BE vuông góc CK , AH // KC
d] Nếu ABC = 60* thì tam giác BAH là tam giác gì?Vì sao ?
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CẢM ƠN.
8
Cho tam giác ABC vuông tại A, trên cạnh BC lấy điểm D sao cho BA=BD. Từ D kẻ đường thẳng vuông góc với BC, cắt AC tại E.
a) Cho AB=5cm, AC=7cm, tính BC?
b) CMR: tam giác ABE= tam giác DBE
c) Gọi F là giao điểm của DE và BA, chứng minh:EF=EC
d) CMR: BE là trung trực của đoạn thẳng AD.
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) Áp dụng định lí pytago cho \(\Delta ABC\):
\(BC^2=AB^2+AC^2\)
\(BC^2=5^2+7^2\)
\(BC^2=25+49\)
\(BC^2=74\)
\(BC=\sqrt{74}\)
\(\Rightarrow BC=\sqrt{74}\)
Chúc bn hk tốt :D