từ điểm B kẻ \(Bz//Cy=>\angle\left(BCy\right)+\angle\left(CBz\right)=180^o\)(góc trong cùng phía)

\(=>\angle\left(CBz\right)=180^o-130^o=50^o\)

\(=>\angle\left(ABz\right)=\angle\left(ABC\right)+\angle\left(CBz\right)=50^o+72^o=122^o\)

\(=>\angle\left(BAx\right)+\angle\left(ABz\right)=180^o\)

mà 2 góc này ở vị trí trong cùng phía

\(=>Ax//Bz=>Ax//Cy\)

kẻ Bz//Ax=>Bz//Cy

\(=>\angle\left(BAx\right)+\angle\left(ABz\right)=180^o\)(tgosc trong cùng phía)

\(=>\angle\left(ABz\right)=180^o-100^o=80^o\)

\(=>\angle\left(CBz\right)=80+40=120^o=\angle\left(BCy\right)\)(so le trong)

Áp dụng t/c dãy tỉ số bằng nhau

\(\dfrac{a}{2013}=\dfrac{b}{2012}=\dfrac{c}{2011}=\dfrac{a-c}{2}=\dfrac{a-b}{1}=\dfrac{b-c}{1}\\ \Rightarrow a-c=2\left(a-b\right)=2\left(b-c\right)\)

\(\Rightarrow H=\dfrac{\left[2\left(a-b\right)\right]^4}{\left(a-b\right)^2\left(a-b\right)^2}=\dfrac{16\left(a-b\right)^4}{\left(a-b\right)^4}=16\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\c+a=2b\\a+b=2c\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=\dfrac{2a}{a}+\dfrac{2b}{b}+\dfrac{2c}{c}=2+2+2=6\)

P=
\(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=\dfrac{a}{b+c}.\left(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\right):\left(\dfrac{a}{b+c}\right)=\left(\dfrac{b+c}{a}.\dfrac{a}{b+c}+\dfrac{c+a}{b}.\dfrac{a}{b+c}+\dfrac{a+b}{c}.\dfrac{a}{b+c}\right):\dfrac{a}{b+c}=\left(\dfrac{b+c}{a}.\dfrac{a}{b+c}+\dfrac{c+a}{b}.\dfrac{b}{c+a}+\dfrac{a+b}{c}.\dfrac{c}{a+b}\right):\dfrac{a}{b+c}=\left(1+1+1\right):\dfrac{a}{b+c}=3.\dfrac{b+c}{a}=\dfrac{3b+3c}{a}\)