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Apollonius theorem states that the sum of the squares of two sides of a triangle is equal to twice the squares of the median on the third side plus half the squares of the third side.
Hence,
AB^2+AC ^2 = 2BD^2+2AD^2
= 2×(1/2×BC)^2+AD^2
=1/2×BC^2+2AD^2
Therefore,
2AB^2+2AC^2 = BC^2+4AD^2---------(1)
similarly,we get, 2AB^2+2BC^2=AC^2+4BE^2---------(2)
2BC^2+2AC^2=AB^2+4CF^2------------(3)
Adding (1),(2) and (3), we get,
4AB^2+4BC^2+4AC^2=AB^2+BC^2+AC^2+
4AD^2+4BE^2+4CF^2
4AB^2-AB^2+4BC^2-BC^2+4AC^2-AC^2 =4(AD^2+BE^2+CF^2)
3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2)
Hope this will help u
Hence,
AB^2+AC ^2 = 2BD^2+2AD^2
= 2×(1/2×BC)^2+AD^2
=1/2×BC^2+2AD^2
Therefore,
2AB^2+2AC^2 = BC^2+4AD^2---------(1)
similarly,we get, 2AB^2+2BC^2=AC^2+4BE^2---------(2)
2BC^2+2AC^2=AB^2+4CF^2------------(3)
Adding (1),(2) and (3), we get,
4AB^2+4BC^2+4AC^2=AB^2+BC^2+AC^2+
4AD^2+4BE^2+4CF^2
4AB^2-AB^2+4BC^2-BC^2+4AC^2-AC^2 =4(AD^2+BE^2+CF^2)
3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2)
Hope this will help u
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