lmaf giúp emmm với ạaaa 50đ luôn ạ

$\begin{array}{l}
33)A + B + C = {180^o} \to \dfrac{{A + C}}{2} = {90^o} - \dfrac{B}{2}\\
\cos \left( {{{180}^o} - \dfrac{B}{2}} \right) =  - \cos \left( {\dfrac{B}{2}} \right) \to {\cos ^2}\left( {{{180}^o} - \dfrac{B}{2}} \right) = {\cos ^2}\dfrac{B}{2}\\
\tan \left( {{{90}^o} - \dfrac{B}{2}} \right) = \tan \dfrac{B}{2}\\
{\cos ^2}\left( {{{180}^o} - \dfrac{B}{2}} \right) - {\cos ^2}\left( {\dfrac{{180 + A + C}}{2}} \right) + \tan \dfrac{B}{2}\tan \dfrac{{A + C}}{2}\\
 = {\cos ^2}\dfrac{B}{2} - {\cos ^2}\left( {{{90}^o} + \dfrac{{A + C}}{2}} \right) + \tan \dfrac{B}{2}\tan \left( {{{90}^o} - \dfrac{B}{2}} \right)\\
 = {\cos ^2}\dfrac{B}{2} - {\cos ^2}\left( {{{90}^o} + {{90}^o} - \dfrac{B}{2}} \right) + \tan \dfrac{B}{2}\cot \dfrac{B}{2}\\
 = {\cos ^2}\dfrac{B}{2} - {\cos ^2}\left( {{{180}^o} - \dfrac{B}{2}} \right) + 1\\
 = {\cos ^2}\dfrac{B}{2} - {\cos ^2}\dfrac{B}{2} + 1 = 1\\
34)\\
Heron:S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \\
S = \dfrac{1}{4}\left( {a + b - c} \right)\left( {a - b + c} \right) = \dfrac{{a + b + c}}{2}.\dfrac{{a - b + c}}{2}\\
 = \dfrac{{a + b + c}}{2}.\dfrac{{a + b + c - 2b}}{2} = p\left( {p - b} \right)\\
 \Leftrightarrow \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)}  = p\left( {p - b} \right)\\
 \Leftrightarrow p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right) = {p^2}{\left( {p - b} \right)^2}\\
 \Leftrightarrow \left( {p - a} \right)\left( {p - c} \right) = p\left( {p - b} \right)\\
 \Leftrightarrow {p^2} - cp - ap + ac = {p^2} - bp\\
 \Leftrightarrow pc + ap - ac = pb\\
 \Leftrightarrow p\left[ {b - \left( {a + c} \right)} \right] + ac = 0\\
 \Leftrightarrow \dfrac{{\left( {a + b + c} \right)\left( {b - a - c} \right)}}{2} + ac = 0\\
 \Leftrightarrow {b^2} - {\left( {a + c} \right)^2} + 2ac = 0\\
 \Leftrightarrow {b^2} = {a^2} + {c^2}
\end{array}$

Suy ra $\Delta ABC$ vuông tại $B$