Question

A circle having centre p circumscribes a triangle ABC and od is perpendicular to BC prove that angle bod is equal to angle a

Answer 1

Solution:

Keep in mind →→Angle subtended by an arc at the center is twice the angle subtended by it any point on the circle.

A Triangle ABC circumscribed in A circle having center O.

Draw OD ⊥ BC.

⊥ from the center to chord bisects the chord.

So, BD= DC

In ΔBDO and ΔCDO

BD = CD →→[Proved above]

BO=OC →→[Radii of circle]

OD is Common.

∠BDO = ∠CDO=90°

ΔBDO ≅ ΔCDO→→→[SAS]

∠BOD = ∠COD→→[CPCT]

∠BOC= 2 ∠BAC→→→Angle subtended by an arc at the center is twice the angle subtended by it any point on the circle.

∠BOD + ∠COD= 2 ∠BAC

2 ∠BOD= 2 ∠BAC

Cancelling 2 from both sides

∠BOD=  ∠BAC→→Hence Proved.