Giải hộ mình bài 5,8,9

Giải thích các bước giải:

Bài 5:

Ta có:

$\widehat{ABC}=180^o-\hat A-\hat C=60^o$

$BD$ là phân giác $\hat B\to \widehat{ABD}=\widehat{DBC}=\dfrac12\widehat{ABC}=30^o$

$\widehat{BDA}=180^o-\hat A-\widehat{ABD}=110^o$

$\widehat{BDC}=180^o-\widehat{BDA}=70^o$

Bài 8:

Ta có:

$\widehat{AKC}=180^o-(\widehat{KAC}+\widehat{KCA})$

$\to \widehat{AKC}=180^o-(\dfrac12\hat A+\dfrac12\hat C)$

$\to \widehat{AKC}=180^o-\dfrac12(\hat A+\hat C)$

$\to \widehat{AKC}=180^o-\dfrac12(180^o-\hat B)$

$\to \widehat{AKC}=180^o-\dfrac12(180^o-\hat 72^o)$

$\to \widehat{AKC}=126^o$

Bài 9:

Ta có:

$\widehat{HNC}=180^o-(\widehat{NHC}+\widehat{NCH}$

$\to \widehat{HNC}= 180^o-(\dfrac12\hat H+\dfrac12\hat C)$

$\to \widehat{HNC}=180^o-\dfrac12(\hat H+\hat C)$

$\to \widehat{HNC}=180^o-\dfrac12(180^o-\hat E)$

$\to \widehat{HNC}=180^o-90^o+\dfrac12\hat E$

$\to \widehat{HNC}=90^o+\dfrac12\hat E$

$\to 123^o=90^o+\dfrac12\hat E$

$\to \dfrac12\hat E=33^o$

$\to \hat E=66^o$