Câu 4:
a: yy'⊥MN
zz'⊥MN
Do đó: yy'//zz'
b: Ta có: \(\hat{xAM}+\hat{BAM}=180^0\) (hai góc kề bù)
=>\(\hat{BAM}=180^0-60^0=120^0\)
Ta có: zz'//yy'
=>\(\hat{BAM}=\hat{zBA}\) (hai góc so le trong)
=>\(\hat{zBA}=120^0\)
c: BH là phân giác của góc zBA
=>\(\hat{zBH}=\hat{ABH}=\frac12\cdot120^0=60^0\)
AH//Bz
=>\(\hat{AHB}=\hat{zBH}\) (hai góc so le trong)
=>\(\hat{AHB}=60^0\)
d: Bt là phân giác của góc x'Bz'
=>\(\hat{x^{\prime}Bt}=\hat{z^{\prime}Bt}=\frac12\cdot120^0=60^0\)
Ta có: \(\hat{x^{\prime}Bt}=\hat{ABH}\left(=60^0\right)\)
mà \(\hat{x^{\prime}Bt}+\hat{ABt}=180^0\) (hai góc kề bù)
nên \(\hat{ABH}+\hat{ABt}=180^0\)
=>BH và Bt là hai tia đối nhau
Câu 5:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019\cdot\left(bk\right)^2+2020b^2}{2019\cdot\left(bk\right)^2-2020b^2}=\frac{b^2\left(2019k^2+2020\right)}{b^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)
\(\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019\cdot\left(dk\right)^2+2020d^2}{2019\cdot\left(dk\right)^2-2020d^2}=\frac{d^2\left(2019k^2+2020\right)}{d^2\left(2019k^2-2020\right)}=\frac{2019k^2+2020}{2019k^2-2020}\)
Do đó: \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\)
