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  • In Fig. 6.54, O is a point in the interior of a triangle ABC, OD perpendicular to BC, OE perpendicular to AC and OF perpendicular to AB. Show that AF ^2 + BD ^2 + CE ^2 equals AE ^2 + CD ^2 + BF ^2

Q8 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.  AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.

  

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best_answer

Join AO, BO, CO

In \triangle AOF, by Pythagoras theorem,

OA^2=OF^2+AF^2..................1

In \triangle BOD, by Pythagoras theorem,

OB^2=OD^2+BD^2..................2

In \triangle COE, by Pythagoras theorem,

OC^2=OE^2+EC^2..................3

Adding equation 1,2,3,we get

OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4

\Rightarrow (OA^2-OE^2)+(OC^2-OD^2)+(OB^2-OF^2)=AF^2+BD^2+EC^2\Rightarrow AE^2+CD^2+BF^2=AF^2+BD^2+EC^2

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seema garhwal

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