In Fig. 6.54, O is a point in the interior of a triangleABC, OD BC, OE AC and OF AB. Show that


(i)


(ii)

Join OA, OB, and OC

(i) Applying Pythagoras theorem in ΔAOF, we obtain


OA2 = OF2 + AF2


Similarly, in ΔBOD,


OB2 = OD2 + BD2


Similarly, in ΔCOE,


OC2 = OE2 + EC2


Adding these equations, we get


OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2


OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + EC2


(ii) From the above given result,


AF2 + BD2 + EC2 = (AO2 – OE2) + (OC2 – OD2) + (OB2 – OF2)


Therefore,


AF2 + BD2 + EC2 = AE2 + CD2 + BF2


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