In triangle abc'E is the mid …
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Sia ? 6 years, 2 months ago
AD is median so,
ar ({tex}\triangle{/tex}ABD) = {tex}\frac{1}{2}{/tex}ar({tex}\triangle{/tex}ABC) ...(i)
BE is median so,
ar ({tex}\triangle{/tex}BED) = {tex}\frac{1}{2}{/tex}ar (ABD) ... (ii)
From (i) & (ii)
ar ({tex}\triangle{/tex}BED) = {tex}\frac{1}{4}{/tex}ar ({tex}\triangle{/tex}ABC)
Hence, Proved
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