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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 OA^2 + OB^2 + OC^2 – OD^2 – OE^2 – OF^2 = AF^2 + BD^2 + CE^2,

Q8 (1)   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
           OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2,

        

Answers (1)
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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

Hence proved

 

 

 

 

 

 

 

 

 

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