Question

[nguyễn ngân hà]

giúp mk 2 bài này vs ạ

Answer 1

[Phong]

3.5:

Xét `\DeltaCID` có:

`\hat{IDC}+\hat{ICD}+\hat{CID}=180^o`

`\hat{IDC}+\hat{ICD}=180^o-\hat{CID}`

`\hat{IDC}+\hat{ICD}=180^o-105^o=75^o`

Mà: `ID,IC` lần lượt là phân giác của `\hat{D},\hat{D}

Suy ra: `\hat{D}=2\hat{IDC},\hat{C}=2\hat{ICD}`

`->1/2\hat{D}+1/2\hat{C}=75^o`

`->\hat{D}+\hat{C}=75^o*2=150^o`

Xét tứ giác `ABCD` có:

`\hat{A}+\hat{B}+\hat{C}+\hat{D}=360^o`

`->\hat{A}+\hat{B}=360^o-(\hat{C}+\hat{D})`

`->\hat{A}+\hat{B}=360^o-150^o`

`->\hat{A}+\hat{B}=210^o`

`->\hat{A}=210^o-\hat{B}`

Mà: `\hat{A}-\hat{B}=30^o`

`->210^o-\hat{B}-\hat{B}=30^o`

`->2\hat{B}=210^o-30^o=180^o`

`->\hat{B}=180^o/2=90^o`

Suy ra: `\hat{A}=210^o-90^o=120^o`

Answer 2

[Nguyễn Lê Phước Thịnh]

3.5:

Xét ΔCID có \(\hat{ICD}+\hat{IDC}+\hat{CID}=180^0\)

=>\(\hat{ICD}+\hat{IDC}=180^0-105^0=75^0\)

=>\(\frac12\left(\hat{BCD}+\hat{CDA}\right)=75^0\)

=>\(\hat{BCD}+\hat{CDA}=75^0\cdot2=150^0\)

Xét tứ giác ABCD có \(\hat{BAD}+\hat{ABC}+\hat{BCD}+\hat{CDA}=360^0\)

=>\(\hat{BAD}+\hat{ABC}=360^0-150^0=210^0\)

mà \(\hat{BAD}-\hat{ABC}=30^0\)

nên \(\hat{BAD}=\frac{210^0+30^0}{2}=120^0;\hat{ABC}=120^0-30^0=90^0\)

3.4:

Xét tứ giác ABCD có \(\hat{BAD}+\hat{ABC}+\hat{BCD}+\hat{ADC}=360^0\)

=>\(\hat{BCD}+\hat{ADC}=360^0-\left(\hat{BAD}+\hat{ABC}\right);\hat{BAD}+\hat{ABC}=360^0-\left(\hat{BCD}+\hat{ADC}\right)\)

Xét ΔEAB có \(\hat{EAB}+\hat{EBA}+\hat{AEB}=180^0\)

=>\(\hat{BEA}+\frac12\left(\hat{BAD}+\hat{ABC}\right)=180^0\)

=>\(\hat{BEA}+\frac12\left\lbrack360^0-\left(\hat{ADC}+\hat{BCD}\right)\right\rbrack=180^0\)

=>\(\hat{BEA}+180^0-\frac12\left(\hat{ADC}+\hat{BCD}\right)=180^0\)

=>\(\hat{BEA}=180^0-180^0+\frac12\left(\hat{ADC}+\hat{BCD}\right)=\frac12\left(\hat{ADC}+\hat{BCD}\right)\)

Vì AE và AF là hai tia phân giác của hai góc kề bù

nên AE⊥AF

Vì BF,BE là hai tia phân giác của hai góc kề bù

nên BF⊥BE

Xét tứ giác FAEB có \(\hat{FAE}+\hat{FBE}+\hat{AFB}+\hat{AEB}=360^0\)

=>\(\hat{AFB}+\hat{AEB}=360^0-90^0-90^0=180^0\)

=>\(\hat{AFB}=180^0-\frac12\left(\hat{ADC}+\hat{BCD}\right)=\frac12\left(360^0-\hat{ADC}-\hat{BCD}\right)=\frac12\cdot\left(\hat{BAD}+\hat{ABC}\right)\)