Lời giải:
\(\widehat{BAD}=\widehat{BAH}+\widehat{HAD}=(90^0-\widehat{BAH})+\frac{\widehat{HAC}}{2}\)
\(=(90^0-\widehat{BAH})+\frac{90^0-\widehat{HCA}}{2}\)
\(=(90^0-\widehat{B})+\frac{90^0-\widehat{C}}{2}=\widehat{C}+\frac{90^0-\widehat{C}}{2}=\frac{90^0+\widehat{C}}{2}(1)\)
Và:
\(\widehat{BDA}=\widehat{DAC}+\widehat{DCA}=\frac{\widehat{HAC}}{2}+\widehat{DCA}=\frac{90^0-\widehat{HCA}}{2}+\widehat{DCA}\)
\(=\frac{90^0-\widehat{C}}{2}+\widehat{C}=\frac{90^0+\widehat{C}}{2}(2)\)
Từ \((1)(2)\Rightarrow \widehat{BAD}=\widehat{BDA}\Rightarrow \triangle ABD\) cân tại $B$
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