Question

D and E are points on the side CA and CB respectively of triangle ABC right angled at C prove that AE²+BD²=AB²+DE²

Answer 1

I think you mean AE​2 + BD2  = AB2 +​ DE2

In triangle ACE,
AC2 ​+ CE2 = AE2 ...(i)    (By Pythagoras Theorem)
In triangle DBC,
DC2 ​+ BC2 = BD2 ...(ii)    (By Pythagoras Theorem)
In triangle ABC,
AC2 + BC2 = AB2 ...(iii)    (By Pythagoras Theorem)
In triangle DEC,
DC2 + CE2 = DE2 ...(iv)     (By Pythagoras Theorem)​

Adding (i) and (ii) we get,
AE2 + BD2 = AC2 + CE2+ DC2+ BC2
AE2 +BD2 = AC2 +BC2 +CE2 + DC2
AE2 + BD2 = AB2 + DE2                                 [ from (iii) and (iv)]
              
                     Hence proved
                  hope this helps :)
mark it as brainilist...
     

Answer 2

Given:              A right triangle ABC, right angled at C. D and E are

                        points on side AC and BC respectively.

To prove:         AE^2+BD^2=AB^2+DE^2

Construction: Join AE, BC and DE

Proof:               In triangle ACE

                        AE^2=AC^2+CE^2___________(I)

                        [By Pythagorean’s theorem]

                        In triangle BCD

                       BD^2=CD^2+BC^2___________(Il)

                       [By Pythagorean’s theorem]

                      Add (l) and (ll) we get,

                      AE^2+BD^2=(AC^2+BC^2) + (CE^2+CD^2)

                      AE^2+BD^2=AB^2+DE^2

                      Hence proved