Question

In an isosceles triangle ABC ,with AB =AC the bisector of angle B and angle C intersect each other at O ,join A to O, show that (1) OB =OC (2) AO bisect angle A .​

Answer 1

1) /_B = /_C

=> /_B ÷ 2 = /_C ÷ 2

=> /_OBC = /_OCB

But they are adjacent angles

therefore, there opposite sides are equal

so, OB = OC................(i)

2) In ∆AOB & ∆AOC,

• OA = OA. [common]
• OB = OC. [From (i)]
• AC = AB. [Given]

Therefore, ∆AOB = ∆AOC [SSS]

So, /_OAB = /_OAC [CPCT]

Therefore, OA is bisector of /_A

Answer 2

Step-by-step explanation:

$\huge\color{Red}{\colorbox{black}{XxItzAdarshxX }}$

Solution:-

Given:-

AB = AC and

the bisectors of B and C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

B = C

½ B = ½ C

⇒ OBC = OCB (Angle bisectors)

∴ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects A.

$\large\bf{\underline\green{❥thαnk \; чσu ♥♥}}$