Question

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that `A E^2+B D^2=A B^2+D E^2`.

Answer 1

In`/_ABC`
`AB^2=AC^2+CB^2-(1)`
In`/_ACE`
`AE^2=AC^2+CE^2-(2)`
In`/_DCE`
`DE^2=DC^2+CE^2(3)`
In`/_DCB`
`BD^2=DC^2+CE^2-(4)`
adding equaiton 2 and 4
`AE^2+BD^2=AC^2+CE^2+DC^2+CE^2`
From equation 1 and 3
`AE^2+BD^2=AB^2+DE^2`