Question

Q.. In an isosceles triangle ABC, with AB=AC,the bisector of angle B and angel C intersect each other at O. Join A to O. show that : (i) OB=OC. (ii) AO bisects angle A..... Note :::: don't dare to spam with my questions otherwise you will be reported...... xd... ​

Answer 1

Answer:

(i) In △ABC, we have

(i) In △ABC, we have AB=AC

(i) In △ABC, we have AB=AC⇒∠C=∠B ∣ Since angles opposite to equal sides are equal

(i) In △ABC, we have AB=AC⇒∠C=∠B ∣ Since angles opposite to equal sides are equal⇒

1/2 ∠B= 1/2∠C

⇒∠OBC=∠OCB

⇒∠OBC=∠OCB⇒∠ABO=∠ACO …(1)

⇒∠OBC=∠OCB⇒∠ABO=∠ACO …(1)⇒OB=OC ∣ Since sides opp. to equal ∠s are equal …(2)

⇒∠OBC=∠OCB⇒∠ABO=∠ACO …(1)⇒OB=OC ∣ Since sides opp. to equal ∠s are equal …(2)(ii) Now, in △ABO and △ACO, we have

⇒∠OBC=∠OCB⇒∠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

⇒∠OBC=∠OCB⇒∠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)

⇒∠OBC=∠OCB⇒∠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)

⇒∠OBC=∠OCB⇒∠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

⇒∠OBC=∠OCB⇒∠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

⇒∠OBC=∠OCB⇒∠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

⇒∠OBC=∠OCB⇒∠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

Answer 2

i) In △ABC, we have

AB=AC

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

⇒

2

1

∠B=

2

1

∠C

⇒∠OBC=∠OCB

⇒∠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

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