Question

in the figure 6.9, OD is a bisector of angle BOC and OD perpendicular of OE. Show that the points A, O, B are collinear.

Answer 1

Step-by-step explanation:

Given In the figure, OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC. To show Points A, O and B are collinear i.e., AOB is a straight line. Proof Since, OD and OE bisect angles ∠AOC and ∠BOC, respectively. So, ∠AOC and ∠COB are forming linear pair.

Answer 2

Answer:

Given In the figure, OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC. To show Points A, O and B are collinear i.e., AOB is a straight line. Proof Since, OD and OE bisect angles ∠AOC and ∠BOC, respectively. So, ∠AOC and ∠COB are forming linear pair

Given:

OD is the bisector of ∠AOC

OE is the bisector of ∠BOC

OD⊥OE

Proof:

From given we have ∠AOD=∠DOC .... (1) and ∠BOE=∠EOC .... (2)

∠DOC+∠EOC=90

o

...... (a)

Substituting from 1 and 2

∠AOD+∠BOE=90

o

...... (b)

(a)+(b)

∠AOD+∠DOC+∠EOC+∠BOE=90

o

+90

o

∠AOD+∠DOC+∠EOC+∠BOE=180

o

∴ ∠AOB=180

0

and A,O and B are Collinear.

∠AOD+∠DOC=∠AOC and ∠BOE+∠EOC=∠BOC

∴ ∠AOC+∠BOC=180

o