Xét `ΔABD` có:

$\widehat{B}$+$\widehat{BAD}$+$\widehat{ADB}$`=180^0`(tổng `3` góc trong tam giác)

`=>90^0+20^0+`$\widehat{ADB}$ `=180^0`

`=>`$\widehat{ADB}$`=180^0-90^0-20^0`

`=>`$\widehat{ADB}$`=70^0`

Ta có:

$\widehat{ADB}$+$\widehat{ADC}$`=180^0` (kề bù)

`=>`$\widehat{ADC}$`=180^0-70^0=110^0`

Xét `ΔACD` có:

$\widehat{ADC}$ + $\widehat{DAC}$+$\widehat{C}$`=180^0` (tổng `3` góc trong tam giác)

`=>110^0 + x+x=180^0`

`=>2x=180^0-110^0`

`=>2x=70^0`

`=>x=70^0 :2`

`x=35^0`

Vậy `x=35^0`

 

Xét `\triangleADB` có :

`\hat{BAD}+\hat{B}+\hat{BDA}=180^o`

`=>\hat{BDA}=180^o-20^o-90^o`

`=>\hat{BDA}=70^o`

Ta có : `\hat{BDA}+\hat{ADC}=180^o`

`=>\hat{ADC}=180^o-70^o`

`=>\hat{ADC}=110^o`

Xét `\triangleADC` có :

`\hat{DAC}+x+x=180^o`

`=>2x+=180^o-110^o`

`=>2x=70^o`

`=>x=(70^o)/2=35^o`