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In an isosceles ∆ with AB=AC the bisectors of angle B and angle C intersect each other at O. Show that (1) OB=OC (2) AO bisects angle A
Answers
[Diagram Attached.]
Given
- ABC is an isosceles triangle.
- AB = AC
- Bisectors of ∠B and ∠C intersect each other at O.
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To Prove
(i) OB = OC
(ii) AO bisects ∠A
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Proof
(i) According to the figure,
In ΔABC,
AB = AC [Given]
∠B = ∠C [Angles opposite to the equal sides of an isosceles triangle are equal as well]
∴
∴ ∠CBO =∠BCO
∴ ∠ABO = ∠ACO - (i)
Using the property of angles and sides of an isosceles triangle, which states that sides opposite to the equal angles are equal as well, we can say that -:
OB = OC
Hence, proved.
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(ii) According to the figure,
In ΔABO and ΔACO,
(S) AB = AC [Given]
(A) ∠ABO = ∠ACO [From (i) ; OC bisects ∠C and OB bisects ∠B]
(S) AO = OA [Common Side]
∴ ΔABO ≅ ΔACO by SAS Congruency.
By C.P.C.T,
∠BAO = ∠OAC
∴ OA bisects ∠A.
Hence proved.
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Answer:
Correct 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:
☯ i) OB = OC
☯ ii) AO bisects ∠A
Given :-
✯ In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O.
Find Out :-
☯ i) OB = OC
☯ ii) AO bisects ∠A
Solution :-
i) OB = OC
➸ AB = AC [given]
➸ ∠ACB = ∠ABC [Angles opposite to equal sides of an isosceles triangle are equal]
➙ ∠ACB = ∠ABC
➙ ∠OCB = ∠OBC [Since OB and OC are the angle bisectors of ∠ABC and ∠ACB]
∴ OB = OC [Sides opposite to equal angles of an isosceles triangle are also equal]
HENCE PROVED :-
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
ii) AO bisects ∠A
➸ In ΔOAB and ΔOAC,
➸ AO = AO [Common]
➸ AB = AC [Given]
➸ OB = OC [We already prove this above :)]
Hence,
➸ ΔOAB ≅ ΔOAC [By SSS congruence rule]
➙ ∠BAO = ∠CAO [CPCT (Corresponding parts of congruent triangles) ]
∴ AO bisects ∠A or AO is the angle bisector of ∠A.
HENCE PROVED :-
[Note :- Please refer that attachment for your figure]