Ta có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tính chất tổng 3 góc trong 1 tam giác)

\(\Rightarrow\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2}=90^o\)

\(\Rightarrow\dfrac{\widehat{B}+\widehat{C}}{2}=90^o-\dfrac{\widehat{A}}{2}\)

\(\Rightarrow\)\(tan\left(\dfrac{\widehat{B}+\widehat{C}}{2}\right)=tan\left(90^o-\widehat{\dfrac{A}{2}}\right)\)

\(\Rightarrow tan\left(\dfrac{\widehat{B}+\widehat{C}}{2}\right)=cot\dfrac{A}{2}\)

Theo tính chất của tam giác, ta có:

\(A+B+C=180^0\)

\(\Rightarrow\dfrac{A+B+C}{2}=90^0\)

\(\Rightarrow\dfrac{B+C}{2}=90^0-\dfrac{A}{2}\)

\(\Rightarrow tan\left(\dfrac{B+C}{2}\right)=tan\left(90^0-\dfrac{A}{2}\right)\)

\(\Rightarrow tan\left(\dfrac{B+C}{2}\right)=cot\left(\dfrac{A}{2}\right)\)

Bài 1: 

a) Xét ΔABD vuông tại D và ΔACE vuông tại E có 

\(\widehat{BAD}\) chung

Do đó: ΔABD∼ΔACE(g-g)

2.

ĐK: \(x\ne0\)

\(10\left(x+\dfrac{1}{x}\right)^2+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x+\dfrac{1}{x}\right)^2+5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x}-x^2-\dfrac{1}{x^2}-2\right)^2=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x+\dfrac{1}{x}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow\left(x-5\right)^2-5=20\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=5\\x-5=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=0\left(l\right)\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=10\)

Đặt hệ trục:

$B(0,0,0),\ C(a,0,0),\ A(0,a,0)$

$\Rightarrow S(0,a,a\sqrt3)$

$M\left(\dfrac a2,\dfrac a2,0\right)$

Vectơ pháp tuyến:

Mặt $(SAB)$:

$\vec{SA}=(0,0,a\sqrt3),\ \vec{SB}=(0,-a,-a\sqrt3)$

$\Rightarrow \vec n_1=\vec{SA}\times\vec{SB}=(1,0,0)$

Mặt $(SBM)$:

$\vec{SB}=(0,-a,-a\sqrt3)$

$\vec{SM}=\left(\dfrac a2,-\dfrac a2,-a\sqrt3\right)$

$\Rightarrow \vec n_2=\vec{SB}\times\vec{SM} =\left(\dfrac{a^2\sqrt3}{2},\dfrac{a^2\sqrt3}{2},-\dfrac{a^2}{2}\right)$

Góc giữa hai mặt phẳng:

$\cos\varphi=\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|} =\dfrac{\sqrt3}{\sqrt7}$

$\sin\varphi=\dfrac{2}{\sqrt7}$

$\cot\varphi=\dfrac{\cos\varphi}{\sin\varphi} =\dfrac{\sqrt3}{2}$

Xét ΔABC có 

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

\(\Leftrightarrow2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\alpha\)

\(\Leftrightarrow\widehat{IBC}+\widehat{ICB}=\dfrac{180^0-\alpha}{2}\)

Xét ΔIBC có

\(\widehat{BTC}+\widehat{IBC}+\widehat{ICB}=180^0\)

\(\Leftrightarrow\widehat{BTC}=180^0-\dfrac{180^0-\alpha}{2}=\dfrac{180^0+\alpha}{2}\)

Bổ sug đề: Cho (O), BD,CE là các dây của (O)

Sửa đề: Chứng minh góc BOE=góc EDB+góc ECB

1/2(góc EDB+góc ECB)

=1/2(1/2sđ cung EB+1/2sđ cung EB)

=1/2sđ cung EB

=1/2*góc BOE

=>góc EDB+góc ECB=góc BOE