In `Delta ABC , D` is the midpoint of BC and `AE bot BC`. If `AC gtAB`, show that `AB^(2)=AD^(2)-BC*DE+(1)/(4)BC^(2)`

Explanation

In `Delta AFB, angle AEB=90^(@)`
`:. AB^(2)=AE^(2)+BE^(2)" "...(i)`
In `Delta AED, angle AED=90^(@)`
` :. AD^(2)=(AE^(2)+DE^(2))`
`rArr AE^(2)=(AD^(2)-DE^(2))`
`:. AB^(2)=(AD^(2)-DE^(2))+ BE^(2)` [ using (i)] ltbr? `= (AD^(2)-DE^(2))+(BD-DE)^(2)`
`=(AD^(2)-DE^(2))+((1)/(2) BC-DE)^(2)`
`=AD^(2)+(1)/(4)BC^(2)-BC*DE`.