D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^{2 }+ BD^{2} = AB^{2} + DE^{2}

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#### Solution

Applying Pythagoras theorem in ΔACE, we obtain

AC^{2} + CE^{2} = AE^{2} ....(i)

Applying Pythagoras theorem in ΔBCD, we get

BC^{2} + CD^{2} = BD^{2} ....(ii)

Using equations (i) and (ii), we get

AC^{2} + CE^{2} + BC^{2} + CD^{2} = AE^{2} + BD^{2} ...(iii)

Applying Pythagoras theorem in ΔCDE, we get

DE^{2} = CD^{2} + CE^{2}

Applying Pythagoras theorem in ΔABC, we get

AB^{2} = AC^{2} + CB^{2}

Putting these values in equation (iii), we get

DE^{2} + AB^{2} = AE^{2} + BD^{2}.

Concept: Right-angled Triangles and Pythagoras Property

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