TÍnh cạnh đáy BC của tam giác cân ABC biết đường cao tương ứng với cạnh đáy bằng 15,6 cm và đường cao tương ứng với cạnh bên là 12 cm

gọi đường cao là AD, CE
$AD^2=AB^2-\frac{BC^2}{4}$
$CE^2=BC^2 -\frac{AB^2}{4}$
=> $ \frac{15BC^2}{16}= \frac{AD^2}{4}+CE^2$
dòng cuối cùng mình ko hiểu –  thanktra 16-06-15 07:51 AM

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