`~rai~`
\(\text{Xét ΔACB và ΔHCA có:}\\\widehat{BAC}=\widehat{AHC}(=90^\circ)\\\widehat{BCA}=\widehat{ACH}(\widehat{C}\quad chung)\\\Rightarrow \Delta ACB\sim\Delta HCA.(g.g).\quad (1)\\\text{Xét ΔACB và ΔHAB có:}\\\widehat{BAC}=\widehat{BHA}(=90^\circ)\\\widehat{ABC}=\widehat{HBA}(\widehat{B}\quad chung)\\\Rightarrow \Delta ACB\sim\Delta HAB(g.g).\quad(2)\\\text{Từ (1) và (2)}\Rightarrow \Delta HCA\sim\Delta HAB(\sim\Delta ACB).\)
