[Solved] In Figure, O is a point in the interior of a triangle. ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
In 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}.
Given, in ΔABC, O is a point in the interior of a triangle.
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC
(i) By Pythagoras theorem in ΔAOF, we have
OA^{2} = OF^{2} + AF^{2}
Similarly, in ΔBOD
OB^{2} = OD^{2} + BD^{2}
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}.
(ii) AF^{2} + BD^{2} + EC^{2} = (OA^{2} – OE^{2}) + (OC^{2} – OD^{2}) + (OB^{2} – OF^{2})
∴ AF^{2} + BD^{2} + CE^{2}
= AE^{2} + CD^{2} + BF^{2}.

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