Question

1. If O is a point within AABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC +CA> OA + OB + OC
1
(iii) OA + OB + OC > ½AB +½ BC + ½CA)​

Answer 1

Step-by-step explanation:

(i) It is given that O is a point within △ ABC Consider △ ABC We know that AB + AC > BC ….. (1) Consider △ OBC We know that OB + OC > BC ….. (2) By subtracting both the equations we get (AB + AC) – (OB + OC) > BC – BC So we get (AB + AC) – (OB + OC) > 0 AB + AC > OB + OC Therefore, it is proved that AB + AC > OB + OC. (ii) We know that AB + AC > OB + OC In the same way we can write AB + BC > OA + OC and AC + BC > OA + OB By adding all the equations we get AB + AC + AB + BC + AC + BC > OB + OC + OA + OC + OA + OB So we get 2 (AB + BC + AC) > 2 (OA + OB + OC) Dividing by 2 both sides AB + BC + AC > OA + OB + OC (iii) Consider △ OAB We know that OA + OB > AB ….. (1) Consider △ OBC We know that OB + OC > BC ….. (2) Consider △ OCA OC + OA > CA …… (3) By adding all the equations OA + OB + OB + OC + OC + OA > AB + BC + CA So we get 2 (OA + OB + OC) > AB + BC + CA Dividing by 2 OA + OB + OC > ½ (AB + BC + CA) Therefore, it is proved that OA + OB + OC > ½ (AB + BC + CA).

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