ABC is a right triangle right-angled at B. Let D and E be any points on AB and BC respectively. Prove that `AE^2+ CD^2= AC^2+ DE^2`

Explanation

As ΔABE is right angled triangle, then,


`⇒ (AE)^2=(AB)^2+(BE)^2` ............(1)


Similarly, ΔDBC is right angled triangle, so,


`⇒ (CD)^2=(BD)^2+(BC)^2` ............(2)


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