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Posted by toanhoclatinhyeucuatoi on 01-08-2018 - 16:16
Cho tam giác ABC, đường cao AH. Giả sử $\frac{1}{AH^{2}}$=$\frac{1}{AB^{2}}$ + $\frac{1}{AC^{2}}$
CMR: $\widehat{B} + \widehat{C} =90^{\circ}$ hoặc |$\widehat{B}$ -$\widehat{C}$| =$90^{\circ}$