Prove that AE2+BD2=AB2+DE2

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Applying Pythagoras theorem in ΔACE, we obtain

Triangles

AC2 + CE2 = AE2    (i)

Applying Pythagoras theorem in triangle BCD, we get

BC2 + CD2 = BD2     (ii)

Using (i) and (ii), we get

AC2 + CE2 + BC2 + CD2 = AE2 + BD2     (iii)

Applying Pythagoras theorem in triangle CDE, we get

DE2 = CD2 + CE2

Applying Pythagoras in triangle ABC, we get

AB2 = AC2 + CB2

Putting these values in (iii), we get

DE2 + AB2 = AE2 + BD2

 

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