Question

In Fig. 16.135, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB=AC.

Answer 1

Given :   O is the centre of the circle, BO is the bisector of ∠ABC.  

 

To Show :  AB = AC.

 

Proof:  

∠ABO = ∠CBO …………..(1)

[BO bisector of ∠ABC]  

OB = OA (Radius)

 Therefore,

∠ABO = ∠DAB ………….. (2)

Angles opposite to equal sides of a triangle are equal

OB = OC (Radius)

Therefore,

∠CBO = ∠OCB ……………(3)

(Angles opposite to equal sides of a triangle are equal)  

From eq (1), (2) and (3)

∠OAB = ∠OCB …………..(4)

In ∆OAB and ∆OCB,  

∠OAB = ∠OCB  

[From eq (4)]

∠OBA = ∠OBC (Given)

OB = OB (Common)

∆OAB ≅ ∆OCB

[By AAS congruence criterion  ]

AB = BC (By c.p.c.t)

Hence, proved

HOPE THIS ANSWER WILL HELP YOU…..

 

Some questions of this chapter :

In Fig. 16.122, O is the centre of the circle. Find ∠BAC.

https://brainly.in/question/15910117

 

ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CO:OP = 3:1. If ar(ΔPBO)=10CM². Find the area of parallelogram ABCD.

https://brainly.in/question/15910119

Answer 2

Step-by-step explanation:

Draw line OD ⊥ AB and line OE ⊥ BC. So now we have the following:

∠ABO = ∠CBO - given as OB is bisector of ∠ABC (A)

∠BDO = ∠BEO = 90°  - by construction(A)

side BO = BO = common side ----- given(A)

Hence Δ DBO ≡ Δ EBO -- by AAS postulate

∴ OD = OE  ---------- by CPCTC 

∴ AB = BC  --- Chords that are equidistant from the center of the circle are equal(or congruent)