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Giải:
Ta có: BD = ED ( gt )
\(\Rightarrow\dfrac{2}{3}BD=\dfrac{2}{3}ED\)
\(\Rightarrow BI=ED\) (1)
\(BD=ED\Rightarrow\dfrac{1}{3}BD=\dfrac{1}{3}ED\Rightarrow ID=DK\)
Lại có: \(DE=\dfrac{1}{3}DE+\dfrac{1}{3}DE+\dfrac{1}{3}DE\)
\(\Rightarrow DE-\dfrac{1}{3}DE=DK+DK\)
\(\Rightarrow\dfrac{2}{3}DE=DK+ID\left(DK=ID\right)\)
\(\Rightarrow KE=IK\) (2)
Từ (1), (2) \(\Rightarrow BI=IK=KE\left(đpcm\right)\)