Question

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

Answer 1

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

Answer 2

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