Question

solve this...........​

Answer 1

Step-by-step explanation:

Given :- In triangle ABC, AB = AC and BO = CO

To Prove :- triangle ABO ≈ triangle ACO

PROOF:-

In ∆ABO & ∆ACO

=} AB = AC. ( GIVEN )

=} OB = OC. ( GIVEN )

=} angle A = angle A ( common)

since , ∆ABO ≈ ∆ACO. proved

Answer 2

Step-by-step explanation:

Given:-

ΔABC where AB = AC, and ΔBOC where BO = CO

Now,

In ΔBOC,

BO = CO

∴∠OBC = ∠OCB ----- 1

[Angles opposite to equal sides are equal]

or

[Properties of Isosceles triangle]

Now,

In ΔABC,

AB = AC

∴∠ABC = ∠ACB ------ 2

[Angles opposite to equal sides are equal]

or

[Properties of Isosceles triangle]

Now,

∠ABC = ∠ABO + ∠OBC

∠ACB = ∠ACO + ∠OCB

From eq.2 we know that,

∠ABO + ∠OBC = ∠ACO + ∠OCB

Again From eq.1 we know that,

∠ABO +∠OBC = ∠ACO + ∠OBC

[As ∠OBC = ∠OCB]

∠ABO + ∠OBC - ∠OBC = ∠ACO

∠ABO + 0 = ∠ACO

∠ABO = ∠ACO

Hence Proved

Hope it helped and believing you understood it........All the best