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# In Fig. 6.54, O is a point in the interior of a triangle. ABC, OD $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ 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 $$\triangle$$ ABC, O is a point in the interior of a triangle.

And OD $$\perp$$ BC, OE $$\perp$$ AC and OF $$\perp$$ AB

Join OA, OB and OC

(i) By Pythagoras theorem in $$\triangle$$ AOF, we have
$$OA^2 = OF^2 + AF^2$$

Similarly, in $$\triangle$$ BOD

$$OB^2 = OD^2 + BD^2$$

Similarly, in $$\triangle$$ COE

$$OC^2 = OE^2 + EC^2$$

$$OA^2 + OB^2 + OC^2 = OF^2 + AF^2 + OD^2 + BD^2 + OE^2 + EC^2$$
$$OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2.$$
$$AF^2 + BD^2 + EC^2 = (OA^2 – OE^2) + (OC^2 – OD^2) + (OB^2 – OF^2)$$
$$AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$$.