Lời giải:
Xét tam giác $ABH$ và $ADH$ có:
\(\widehat{BAH}=\widehat{DAH}\) (gt)
\(\widehat{AHB}=\widehat{AHD}=90^0\)
\(AH\) chung
\(\Rightarrow \triangle ABH=\triangle ADH(g.c.g)\Rightarrow BH=DH\)
\(\Rightarrow DH=\frac{1}{2}BD=\frac{1}{2}DC(1)\)
Kẻ DK\perp AC$ ($K\in AC$)
Xét tam giác $AHD$ và $AKD$ có:
\(\widehat{AHD}=\widehat{AKD}=90^0\)
\(\widehat{HAD}=\widehat{DAC}=\widehat{KAD}\)
$AD$ chung
\(\Rightarrow \triangle AHD=\triangle AKD(g.c.g)\)
\(\Rightarrow DH=DK(2)\)
Từ \((1);(2)\Rightarrow DK=\frac{1}{2}DC\)
Xét tam giác vuông $DKC$ có : \(\sin \widehat{C}=\frac{DK}{DC}=\frac{1}{2}\Rightarrow \widehat{C}=30^0\)
\(\widehat{HAC}=90^0-\widehat{C}=60^0\). Mà \(\widehat{HAC}=\frac{2}{3}\widehat{BAC}\Rightarrow \widehat{BAC}=90^0\)
Vậy:
\(\widehat{ABH}=\widehat{ABC}=180^0-(\widehat{BAC}+\widehat{C})=180^0-(90^0+30^0)=60^0\)
Vậy \(\widehat{ABH}=60^0\)
