Question
In $\triangle\text{ABC},$ AD is median. Prove that $AB^2 + AC^2 = 2AD^2 + 2DC^2.$
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Answer
Given: In $\triangle\text{ABC},$ AD is the median of BC
To prove: $AB^2 + AC^2 = 2AD^2 + 2DC^2$
Construction: Draw $\text{AE}\perp\text{BC}$
Proof: In $\triangle\text{ABE},$
$AB^2 = AE^2 + BE^2 ....(i) $ (Pythagoras Theorem)
Similarly in right $\triangle\text{ACE},$
$AC^2 = AE^2 + EC^2 .....(ii)$
and in right $\triangle\text{AED},$
$AD^2 = AE^2 + ED^2 ....(iii)$
Adding (i) and (ii)
$AB^2 + AC^2 = AE^2 + BE^2 + AE^2 + EC^2$
$= 2AE^2 + (BD - ED)^2 + (DC + ED)^2$
$= 2AE^2 + BD^2 + ED^2 - 2BD \times ED + DC^2 + ED^2 + 2BC \times ED$
$= 2AE^2 + BD^2 + 2ED^2 + BD^2 + 2BD \times ED - 2BD \times ED $($\because$ D is mid points)
$= 2AE^2 + 2BD^2 + 2ED^2$
$= 2(AE^2 + ED^2) + 2DC^2$ ($\because$ DC = BD)
$= 2AD^2 + 2DC^2$ {From (iii)}
Hence proved.
To prove: $AB^2 + AC^2 = 2AD^2 + 2DC^2$
Construction: Draw $\text{AE}\perp\text{BC}$
Proof: In $\triangle\text{ABE},$
$AB^2 = AE^2 + BE^2 ....(i) $ (Pythagoras Theorem)
Similarly in right $\triangle\text{ACE},$
$AC^2 = AE^2 + EC^2 .....(ii)$
and in right $\triangle\text{AED},$
$AD^2 = AE^2 + ED^2 ....(iii)$
Adding (i) and (ii)
$AB^2 + AC^2 = AE^2 + BE^2 + AE^2 + EC^2$
$= 2AE^2 + (BD - ED)^2 + (DC + ED)^2$
$= 2AE^2 + BD^2 + ED^2 - 2BD \times ED + DC^2 + ED^2 + 2BC \times ED$
$= 2AE^2 + BD^2 + 2ED^2 + BD^2 + 2BD \times ED - 2BD \times ED $($\because$ D is mid points)
$= 2AE^2 + 2BD^2 + 2ED^2$
$= 2(AE^2 + ED^2) + 2DC^2$ ($\because$ DC = BD)
$= 2AD^2 + 2DC^2$ {From (iii)}
Hence proved.
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