Question

A point O is in the interior of
Δ ABC .Prove that 2(OA +OB +OC )> AB +BC +AC

Answer 1

Answer:

your answer is

Step-by-step explanation:

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABD

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCOD

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we get

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)Similarly

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)SimilarlyAB+BC>OA+OC.....(4)

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)SimilarlyAB+BC>OA+OC.....(4)And

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)SimilarlyAB+BC>OA+OC.....(4)AndAC+BC>OA+OB.....(5)

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)SimilarlyAB+BC>OA+OC.....(4)AndAC+BC>OA+OB.....(5)Adding (3),(4)and (5) we get

Let O be a inner point of a triangle ABC. We are to prove that OA+OB+OC<AB+BC+CA*Construction"BO is produced to intersect AB at D.For ΔABDAB+AD>BD⇒AB+AD>BO+OD.....[1]For ΔCODCD+OD>OC......[2]Adding [1] and [2] we getAB+AD+CD+OD>BO+OD+OC⇒AB+AC+OD>BO+OD+OC⇒AB+AC>OB+OC.....(3)SimilarlyAB+BC>OA+OC.....(4)AndAC+BC>OA+OB.....(5)Adding (3),(4)and (5) we get2(AC+BC+

Answer 2

Answer:

oa + ob > ab

ob + oc > bc

oa + oc > ac

add them

2(oa + ob + oc ) > ab + bc + ca