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

In the given Fig, O is a point in the interior of a triangle ABC, OD ⊥  BC,OE ⊥ AC and OF ⊥ AB. Show that(i) OA2 + OB2 + OC2 -

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

1 Answer

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

Triangles Exercise 6.5 Answer 8

(i) By Pythagoras theorem in ΔAOF, we have

OA2 = OF2 + AF2

Similarly, in ΔBOD

OB2 = OD2 + BD2

Similarly, in ΔCOE

OC2 = OE2 + EC2

Adding these equations,

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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