Question

5. In triangle ABC, if 'O' is the Circumcentre and 'H' is the Orthocentre,
then show that
i) OA+OB+OC = OH ii) HA+HB + HC = 2HO​

Answer 1

To prove:

i) OA + OB + OC = OH    II) HA + HB + HC  = 2HO

          Given,

                  O = Circumcentre of the triangle

                  H = Orthocentre of the triangle

           Let D be the midpoint BC

     (I)      From figure,

                     OD = (OB + OC)/2    ----------(1)

                     OA + OB + OC = OA + 2(OD)              [FROM (1)]

                     OA + OB + OC = OA + AH

                      OA + OB + OC = OH

      (II)    Consider,

                        HA + HB + HC = (OA - OH) + (OB - OH) + (OC - OH)

                                                  = OA + OB + OC - 3OH

                                                  = OH - 3OH

                                                  = -2OH

                                                  = 2HO

Note: I used bold letters for vector representation.

Answer 2

Answer:

we know that

HG=2GO where G is centroid of triangle

let a point D, between B and C

OD=(OB+OC)/2

OA+OB+OC=OA+2OD

we know that G divide The point A and midpoint

opposite side in ratio 2 :1

OG=

3

OA+2OD

OA+OB+OC=30G=20G+OG

=HG+OG

OA+OB+OC=HO