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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.

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Solution: In ∆ ACE; AE2 = AC2 + CE2 ……… (1)

In ∆ DCB; BD2 = DC2 + CB2 ……… (2)

In ∆ ACB; AB2 = AC2 + CB2 ……… (3)

In ∆ DCE; DE2 = DC2 + CE2 ……… (4)

Adding equations (1) and (2), we get;

AE2 + BD2 = AC2 + CE2 + DC2 + CB2 …….. (5)

Adding equations (3) and (4), we get;

AB2 + DE2 = AC2 + CB2 + DC2 + CE2 ………. (6)

On comparing the RHS of equations (5) and(6), we get;

AE2 + BD2 = AB2 + DE2 proved

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