In given fig, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:

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

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

Asked by Pragya Singh | 1 year ago |  79

##### Solution :-

Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA2 = OF2 + AF2

Similarly, in ΔBOD

OB2 = OD2 + BD2

Similarly, in ΔCOE

OC2 = OE2 + EC2

OA2 + OB2 + OC2 = OF2 + AF2 +

OD2 + BD2 + OE+ EC2

OA2 + OB2 + OC2 – OD2 – OE2 –

OF2 = AF2 + BD2 + CE2.

(ii) AF2 + BD2 + EC2 = (OA2 – OE2) +

(OC2 – OD2) + (OB2 – OF2)

AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

Answered by Abhisek | 1 year ago

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