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ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB. Show that:

\( (i) OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2\) ,

\( (ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2\).

Answer :

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

\(OC^2 = OE^2 + EC^2\)

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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