In the figure, O is a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB.

Question :

In the figure, O is a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB. Show thattriangles

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

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

Solution :

Join AO, BO and CO as shown in fig.

(i)  In right \(\triangle\)s OFA, ODB and OEC, we have

\({OA}^2\) = \({AF}^2\) + \({OF}^2\)

\({OB}^2\) = \({BD}^2\) + \({OD}^2\)

and \({OC}^2\) = \({CE}^2\) + \({OE}^2\)

Adding all these, we get

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

or   \({OA}^2\) + \({OB}^2\) + \({OC}^2\) – \({OD}^2\) – \({OE}^2\) – \({OF}^2\) = \({AF}^2\) + \({BD}^2\) + \({CE}^2\)

(ii)  In right \(\triangle\)s ODB and ODC, we have :

\({OB}^2\) = \({BD}^2\) + \({OD}^2\)  and  \({OC}^2\) = \({OD}^2\) + \({CD}^2\)

or  \({OB}^2\) – \({OC}^2\) = \({BD}^2\) – \({CD}^2\)            ……….(1)

Similarly, we have :

\({OC}^2\) – \({OA}^2\) = \({CE}^2\) – \({AE}^2\)               ………..(2)

\({OA}^2\) – \({OB}^2\) = \({AF}^2\) + \({BF}^2\)              ………..(3)

Adding equations (1), (2) and (3), we get

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

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