Question

in an isosceles triangle ABC with AB=AC, the bisectors of ∠B and ∠C intersect each other at O . join A to O. show that:

|)OB=OC
||)AO bisects ∠A

Answer 1

$\huge{\underline{\bf\pink{Answer :-}}}$

Given:

ABC is isosceles triangle ABC with AB=AC.

To Find:

|)OB=OC

||)AO bisects ∠A

Solution:

(i) In △ABC, we have AB=AC

⇒∠C=∠B (Since angles opposite to equal sides are equal)

⇒1/2 ∠B= 1/2= ∠C

⇒∠ABO=∠ACO    …(1)

⇒OB=OC ∣ Since sides opp. to equal ∠s are equal …(2)

(ii) Now, in △ABO and △ACO, we \: have 

AB=AC ∣ Given

∠ABO=∠ACO ∣ From (1)

OB=OC ∣ From (2)

∴ By SAS criterion of congruence, we have

△ABO≅△ACO

⇒∠BAO=∠CAO ∣ Since corresponding parts of congruent triangles are equal.

⇒ AO bisects ∠A

$\huge\fcolorbox{black}{red}{HENCE PROVED}$

Answer 2

Step-by-step explanation:

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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.

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