Question

In a `triangleABC` the angles at B and C are acute. If BE and CF be drawn perpendiculars on AC and AB respectivel, prove that
`BC^(2) = Abxx BF + AC xx EC`

Answer 1

In right `triangleBFC`
`BC^(2) = BF^(2) + FC^(2)` ( by Pythagores theorem) …(1)
In right `triangleBEC`
`BC^(2) = BE^(2)+EC^(2)`
( by Pythagores theorem) …(2)
Adding (1) and (2) , we get
`2BC^(2)= BF^(2)+FC^(2)+BE^(2)+EC^(2)`
`(AB-AF)^(2)+AC^(2)-AF^(2)+AB^(2)-AE^(2)+(AC-AE) ^(2)` ( by pythagoras theorem)
`= AB^(2)+AF^(2)-2AB.AF-AC^(2)-AF^(2)+AB^(2)-AE^(2)+AC^(2)=AE^(2)-2AC.AE`
`= 2AB^(2) +2AC^(2)-2AB.AF-2AC.AE`
`= 2AB^(2)+2AC^(2)-2AB(AB-BF)-2AC. (AC-EC)`
`Rightarrow " " 2BC^(2) = 2AB xx BF + 2AC xx EC`
`Rightarrow " " BC^(2)= ABxxBF+AC xx EC`