The perpendicular AD on the base BC of a `triangleABC` intersects BC at D, so that DB=3CD. Prove that `2AB^(2)= 2AC^(2)+BC^(2)`

Explanation

Given : A `triangleABC` in which `AB bot BC and DB=3CD`
To prove: `2AB^(2)= 2AC^(2)+BC^(2)`
Proof: since DB=3CD `Rightarrow (DB)/(CD)=3/1`
` DB=3x Rightarrow CD=x`
`(DB)/(BC)= (3x)/(4x)= 3/4 Rightarrow DB = 3/4 BC`
`and (DC)/(BC)= x/(4x)= 1/4 Rightarrow DC = 1/4 BC`
Now , by Pythagoras theorem,
`AB^(2)=AD^(2)| BD^(2)`
`= (AC^(2)- 1/16BC^(2) + 9/16BC^(2)`
`AC^(2)+ 8/16BC^(2)= AC^(2)+1/2BC^(2)`
`2AB^(2)= 2AC^(2) +BC^(2)` Hence proved.