Question

Solve it fast please ​

Answer 1

Answer:

From the given figure, in ∆ ABC,

AB = AC [given data]

∴ ∠ABC = ∠ACB ….. [∵ angles opposite to equal sides are equal]

Also, OB and OC are given as the angle bisectors of angle ABC and angle ACB respectively and BC is produced to D. Therefore,

∠OBA = ∠OBC = ½ ∠ABC ….. (i)

And,  

∠OCB = ∠OCA = ½ ∠ACB ……. (ii)

From (i) & (ii), we get

∠OBC = ∠OCB …… (iii)

Now,  

∠OCD = ∠BOC + ∠OBC ….. [∵ external angle is equal to sum of two remote internal angles]

⇒ ∠OCA + ∠ACD = ∠BOC + ∠OBC

⇒ ∠OCB + ∠ACD = ∠BOC + ∠OBC ….. [from (ii) ∠OCA = ∠OCB]

⇒ ∠OBC + ∠ACD = ∠BOC + ∠OBC ….. [from (iii) ∠OBC = ∠OCB]

⇒ ∠BOC = ∠ACD  

Hence proved

Answer 2

In the given triangle AB = AC, prove that   Angle BOC =  Angle  ACD

Solution :

Given That AB = AC, i.e sides of an isosceles triangle are equal.

By Property of isosceles triangle, the angles opposite to equal sides are also equal.

→Angle ABC = Angle ACB

As Line OB is the angular bisector of Angle ABC, it implies that

Angle ABO =  Angle OBC  and Angle ACO = Angle OCB , which we can assume to be = x

By property of triangle , the sum of all angles = 180 ®,

Hence , the sum of angles of ∆ BOC = 180® i.e  Angle BOC + Angle OCB + Angle CBO = 180®

→Angle BOC +x + x = 180®

→Angle BOC = 180®-2x   ……..Equation 1

Consider the Line BCD :

By property of a straight line, the sum of angles on a horizontal line = 180®

→Angle OBC + Angle OCA + Angle ACD = 180®

→Angle ACD +x+x= 180®

→Angle ACD = 180®-2x………………..Equation 2

From Equation 1 and Equation 2 , it is thus proved that Angle BOC = Angle ACD