cho tam giác ABC có B<90,B=2C. Kẻ đường cao AH.Trên tia đối BA lấy E sao cho BE=BH.Đường thẳng HE cắt AC tại D.Lấy B' sao cho hlaf trung điểm của BB'.Cm :Nếu tam giác ABC vuông thì DE^2=BC^2-AB^2
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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,` vì Tam giác `ABC` có \(\widehat{A}=110^0\)
`=>` Tam giác `ABC` là tam giác tù.
`b,` Cạnh đối diện của \(\widehat{A}\) là cạnh `BC`
`=>` Cạnh lớn nhất của Tam giác `ABC` là cạnh `BC`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Tam giác ABC nhọn:
b) Tam giác ABC vuông tại A:
c) Tam giác ABC có góc A tù:
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\widehat{A}+\widehat{B}+\widehat{C}=180^0$
$\Rightarrow \widehat{A}=180^0-\widehat{B}-\widehat{C}=180^0-45^0-45^0=90^0$
$\Rightarrow$ tam giác $ABC$ là tam giác vuông tại $A$. Mà $\widehat{B}=\widehat{C}$ nên $ABC$ là tam giác vuông cân ở A
![](https://rs.olm.vn/images/avt/0.png?1311)
ΔABC~ΔA'B'C'
=>\(\dfrac{AB}{A'B'}=\dfrac{AC}{A'C'}=\dfrac{BC}{B'C'}\)
=>\(\dfrac{AB}{9}=\dfrac{AC}{12}=\dfrac{BC}{15}\)
=>\(\dfrac{AB}{3}=\dfrac{AC}{4}=\dfrac{BC}{5}\)
=>AB là cạnh nhỏ nhất trong ΔABC
Theo đề, ta có: AB=3cm
=>\(\dfrac{AC}{4}=\dfrac{BC}{5}=\dfrac{3}{3}=1\)
=>\(AC=4\cdot1=4\left(cm\right);BC=5\cdot1=5\left(cm\right)\)