In Fig. O is the centre of the circle, Bo is the bisector of ∠ABC. S

1.	In Fig. O is the centre of the circle, Bo is the bisector of ∠ABC. Show that AB = AC
Answer» Given that, BO is the bisector of ∠ABC To prove : AB = BC Proof : ∠ABO = ∠CBO (BO bisector of ∠ABC) (i) OB = OA (Radii) Therefore, ∠ABO = ∠DAB (Opposite angle to equal sides are equal) (ii) OB = OC (Radii) Therefore, ∠CBO = ∠OCB (Opposite angles to equal sides are equal) (iii) Compare (i), (ii) and (iii) ∠OAB = ∠OCB (iv) In triangle OAB and OCB, we have ∠OAB = ∠OCB [From (iv)] ∠OBA = ∠OBC (Given) OB = OB (Common) By AAS congruence rule, Δ OAB ≅ Δ OCB AB = BC (By c.p.c.t) Hence, proved

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

Answer»

Given that,

BO is the bisector of ∠ABC

To prove :

AB = BC

Proof :

∠ABO = ∠CBO

(BO bisector of ∠ABC)

(i) OB = OA (Radii)

Therefore,

∠ABO = ∠DAB

(Opposite angle to equal sides are equal)

(ii) OB = OC (Radii)

Therefore,

∠CBO = ∠OCB

(Opposite angles to equal sides are equal)

(iii) Compare (i), (ii) and (iii)

∠OAB = ∠OCB

(iv) In triangle OAB and OCB, we have

∠OAB = ∠OCB [From (iv)]

∠OBA = ∠OBC (Given)

OB = OB (Common)

By AAS congruence rule,

Δ OAB ≅ Δ OCB

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

Hence, proved

Discussion

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