Bài 4: Hai đường thẳng song song

Bài 1:
a. $AB\perp AC, CD\perp AC$

$\Rightarrow AB\parallel CD$

b.Áp dụng định lý về tổng 3 góc trong tam giác ta có:

$\widehat{BEA}=90^0-\widehat{ABE}=90^0-30^0=60^0$

$\widehat{DEC}=90^0-\widehat{EDC}=90^0-60^0=30^0$

$\Rightarrow \widehat{BEA}+\widehat{DEC}=60^0+30^0=90^0$

$\widehat{BED}=180^0-(\widehat{BEA}+\widehat{DEC})$

$=180^0-90^0=90^0$

$\Rightarrow BE\perp ED$

(đpcm)

Bài 2:
Áp dụng định lý về tổng 3 góc trong tam giác ta có:

a.

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

$40^0+70^0+x=180^0$

$x=180^0-110^0=70^0$

b. 

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

$50^0+x+x=180^0$

$2x=130^0$

$x=65^0$

c.

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

$90^0+30^0+x=180^0$

$x=180^0-120^0=60^0$

$y=180^0-\widehat{DHC}=\widehat{DHC}+\widehat{DCH}$

$=90^0+30^0=120^0$

Bài 3:

Ta thấy:

$\widehat{ADC}=\widehat{B}+\widehat{BCD}=\widehat{B}+\frac{1}{2}\widehat{C}$
$\widehat{BDC}=\widehat{A}+\widehat{ACD}=\widehat{A}+\frac{1}{2}\widehat{C}$

$\Rightarrow \widehat{BDC}-\widehat{ADC}=\widehat{A}-\widehat{B}=30^0$

Mà: $\widehat{BDC}+\widehat{ADC}=180^0$

Do đó:

$\widehat{BDC}=(180^0+30^0):2=105^0$

$\widehat{ADC}=(180^0-30^0):2=75^0$

Xét tam giác ABC có \(\widehat{BAC}+\widehat{ACB}+\widehat{B}=180\Rightarrow\widehat{BAD}+\widehat{DAC}+\widehat{ACD}+\widehat{BAD}+\widehat{B}=180\) độ

hay \(20+\widehat{DAC}+25+\widehat{ACD}+50=180\) độ

\(\Rightarrow\widehat{DAC}+\widehat{ACD}=85\) độ

Xét tam giác ADC có \(\widehat{DAC}+\widehat{ACD}+\widehat{ADC}=180\Rightarrow\widehat{ADC}=180-85=95\) độ

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