Cho tam giác ABC có góc A vuông, AB = AC

$$\begin{gathered} \mathrm{cho}∆\mathrm{ABC} \mathrm{có} \angle \mathrm{A} =90° \mathrm{và} \mathrm{AB}=\mathrm{AC} . \mathrm{gọi} \mathrm{K} \mathrm{là} \mathrm{trung} \mathrm{điểm} \mathrm{của} \mathrm{BCA} \\ \mathrm{a}, \mathrm{cm}:△\mathrm{AKB}=△\mathrm{AKC} \\ \mathrm{b},\mathrm{cm}:\mathrm{AK}\perp \mathrm{BC} \\ \mathrm{c},\mathrm{từ} \mathrm{C} \mathrm{vẽ} \mathrm{đường} \mathrm{vuông} \mathrm{góc} \mathrm{với} \mathrm{BC} \mathrm{cắt} \mathrm{đưởng} \mathrm{thảng} \mathrm{AB} \mathrm{tại} \mathrm{E}. \mathrm{Cm}:\mathrm{EC}//\mathrm{AK} \end{gathered}$$