In a quadrilateral ABCD, `angleA+angleD=90^(@)`. Prove that
`AC^(2)+BD^(2)=AD^(2)+BC^(2)`

Explanation

Given Quadrilateral ABCD, in which `angleA+angleD=90^(@)`
To prove `AC^(2)+BD^(2)=AD^(2)+BC^(2)`
Construct Produce AB and CD to meet at E.
Also, join AC and BD.
Proof In `DeltaAED, angleA+angleD=90^(@)` [given]
`therefore angleE=180^(@)-(angleA+angleD)=90^(@)`
[`because` sum of angles of a triangle=`180^(@)`]
Then by Pythagoras theroem, `AD^(2)=AE^(2)+DE^(2)`
In `DeltaBEC`, by pythagoras theorem, `BC^(2)=BE^(2)+EF^(2)`
On adding both equations, we get
`AD^(2)+BC^(2)=AE^(2)+DE^(2)+BE^(2)+CE^(2)`..(i)
In `DeltaAEC`, by Pythagoras therom,
`AC^(2)=AE^(2)+CE^(2)`
and in `DeltaBED`, by pythagoras theorem
`BD^(2)=BE^(2)+DE^(2)`
On adding both equation weget
`AC^(2)+BD^(2)=AE^(2)+CE^(2)+BE^(2)+DE^(2)`...(ii)
From Eqs. (i) and (ii),
`AC^(2)+BD^(2)=AD^(2)+BC^(2)` Hence proved