Giải thích các bước giải:

Xét $\Delta BAH,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}(=90^o)$

$\to \Delta HBA\sim\Delta ABC(g.g)$

$\to \dfrac{AB}{BC}=\dfrac{HB}{AB}$

$\to BA^2=BH.BC$

Vì $A, D$ đối xứng qua $B$

$\to B$ là trung điểm $AD$

$\to BA=BD$

$\to BD^2=BH.BC$

$\to \dfrac{BD}{BH}=\dfrac{BC}{BD}$

$\to \Delta BHD\sim\Delta BDC(c.g.c)$

$\to \widehat{BHD}=\widehat{BDC}$

$\to \widehat{AHD}+\widehat{ACD}=(\widehat{AHB}+\widehat{BHD})+\widehat{ACD}=(90^o+\widehat{BDC})+\widehat{ACD}=90^o+\widehat{BDC}+\widehat{ACD}=90^o+\widehat{ADC}+\widehat{ACD}=90^o+90^o=180^o$