Cho tam giác ABC có 2 góc B;C nhọn .Vẽ phía ngoài tam giác ABC các tam giác vuông cân ABD(cân tại B) và ACE (cân tại C) vẽ DI và EK vuông góc với BC (I;K thuộc BC ) Chứng minh:
a)BI=CK
b)BC=ID+EK
K
Khách
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Những câu hỏi liên quan
21 tháng 12 2021
bài 2:
ta có: AB<AC<BC(Vì 3cm<4cm<5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
15 tháng 2 2022
bài 2:
ta có: AB <AC <BC (Vì 3cm <4cm <5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
HT mik làm giống bạn Dương Mạnh Quyết
NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 4
câu 5: Gọi M là giao điểm của AD và BC
Xét ΔBAD có \(\hat{BDM}\) là góc ngoài tại đỉnh D
nên \(\hat{BDM}=\hat{DAB}+\hat{DBA}\)
=>\(\hat{BDM}>\hat{BAD}=\hat{BAM}\) (2)
Xét ΔDAC có \(\hat{MDC}\) là góc ngoài tại đỉnh D
nên \(\hat{MDC}=\hat{DAC}+\hat{DCA}>\hat{DAC}\) (1)
Từ (1),(2) suy ra \(\hat{BDM}+\hat{MDC}>\hat{BAD}+\hat{CAD}\)
=>\(\hat{BDC}>\hat{BAC}\)
Câu 3:
Theo đề, ta có: \(\hat{A}=\hat{B}+25^0;\hat{C}=\hat{B}+35^0\)
Xét ΔBAC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{B}+\hat{B}+25^0+\hat{B}+35^0=180^0\)
=>\(3\cdot\hat{B}=180^0-60^0=120^0\)
=>\(\hat{B}=\frac{120^0}{3}=40^0\)
=>\(\hat{C}=40^0+35^0=75^0\)
Bài 2:
Theo đề, ta có: \(\hat{B}=\hat{A}+24^0;\hat{C}=\hat{A}-30^0\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{A}+\hat{A}+24^0+\hat{A}-30^0=180^0\)
=>\(3\cdot\hat{A}=180^0+30^0-24^0=186^0\)
=>\(\hat{A}=62^0\)
=>\(\hat{C}=62^0-30^0=32^0\)
Câu 1: Theo đề, ta có: \(\hat{B}=\hat{A}+15^0;\hat{C}=\hat{A}+45^0\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{A}+\hat{A}+15^0+\hat{A}+45^0=180^0\)
=>\(3\cdot\hat{A}=180^0-60^0=120^0\)
=>\(\hat{A}=40^0\)
\(\hat{B}=40^0+15^0=55^0\)
22 tháng 2 2020
Câu hỏi của Nguyễn Vũ Thu Hương - Toán lớp 7 - Học toán với OnlineMath
15 tháng 6 2022
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
NN
26 tháng 3 2020
2. \(\Delta ABC\)có AB=AC \(\Rightarrow\Delta ABC\)cân.
AD là phân giác \(\Delta ABC\)mà \(\Delta ABC\)cân.
\(\Rightarrow AD\)l là đường trung trực \(\Delta ABC\)..
\(\Rightarrow AD\)là đường cao \(\Delta ABC\)..
\(\Leftrightarrow AD\perp BC\).