In the figure, given below, AB = AC. Prove that : ∠BOC = ∠ACD.

Question

KnowledgeBoat · Accepted Answer

In △ ABC, ⇒ AB = AC (Given) ∴ ∠C = ∠B = x (let) From figure, OB and OC bisects angle ∠B and ∠C. ∴ ∠OBC = and ∠OCB = . In △ BOC, By angle sum property of triangle, ⇒ ∠OBC + ∠OCB + ∠BOC = 180° ⇒ + ∠BOC = 180° ⇒ x + ∠BOC = 180° ⇒ ∠BOC = 180° - x. From figure, ⇒ ∠OCA = (As OC is bisector of ∠C) Since, BCD is a straight line. ∴ ∠OCB + ∠OCA + ∠ACD = 180° ⇒ + ∠ACD = 180° ⇒ x + ∠ACD = 180° ⇒ ∠ACD = 180° - x. ∴ ∠BOC = ∠ACD. Hence, proved that ∠BOC = ∠ACD.

Mathematics

In the figure, given below, AB = AC. Prove that : ∠BOC = ∠ACD.

Triangles

Related Questions

In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate :
(i) ∠CDE
(ii) ∠DCE

Calculate x :
(i)
(ii)

In the figure given below, LM = LN; angle PLN = 110°. Calculate :
(i) ∠LMN
(ii) ∠MLN

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°. Find :
(i) ∠DCB
(ii) ∠CBD

Mathematics

In the figure, given below, AB = AC. Prove that : ∠BOC = ∠ACD.

Triangles

Related Questions

In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate :(i) ∠CDE(ii) ∠DCE

Calculate x :(i)(ii)

In the figure given below, LM = LN; angle PLN = 110°. Calculate :(i) ∠LMN(ii) ∠MLN

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°. Find :(i) ∠DCB(ii) ∠CBD

In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate :
(i) ∠CDE
(ii) ∠DCE

Calculate x :
(i)
(ii)

In the figure given below, LM = LN; angle PLN = 110°. Calculate :
(i) ∠LMN
(ii) ∠MLN

An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°. Find :
(i) ∠DCB
(ii) ∠CBD