In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA^{2} + OB^{2} + OC^{2} − OD^{2} − OE^{2} − OF^{2} = AF^{2} + BD^{2} + CE^{2}

(ii) AF^{2} + BD^{2} + CE^{2 }= AE^{2} + CD^{2} + BF^{2}

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#### Solution

(i) Applying Pythagoras theorem in ΔAOF, we obtain

OA

^{2}= OF^{2}+ AF^{2}OA^{2} = OF^{2} + AF^{2}

Similarly, in ΔBOD

OB2 = OD2 + BD2

Similarly, in ΔCOE

OC^{2} = OE^{2} + EC^{2}

Adding these equations,

OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2} + EC^{2}

OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}.

From above result

AF^{2} + BD^{2} + EC^{2} = (OA^{2} - OE^{2}) + (OC^{2} - OD^{2}) + (OB^{2} - OF^{2})

∴ AF^{2} + BD^{2} + CE2 = AE^{2} + CD^{2} + BF^{2}.

Concept: Right-angled Triangles and Pythagoras Property

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