Question

If o is the circumcentre of a triangle ABC and OD perpendicular then prove thatangle BOD=BAC

Answer 1

Answer:

The solution is explained step-wise below :

Step-by-step explanation:

For better explanation of the solution see the attached figure :

O is the circumcenter of ΔABC and OD ⊥ BC

∴ O is the point of intersection of perpendicular bisectors of the sides of ΔABC.

⇒ D is the midpoint of BC

⇒ BD = DC

In ΔOBD and ΔOCD,

OB = OC (Radius of circle)

OD = OD (Common)

BD = DC (Proved)

ΔOBD ≅ ΔOCD (SSS congruency criterion)

⇒ ∠BOD = ∠COD (By CPCT )

$\implies \angle BOD =\angle COD=\frac{1}{2}\angle BOC.....(1)$

We know that, angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle

⇒ ∠BOC = 2∠BAC

⇒ 2∠BOD = 2∠BAC  (From equation (1))

⇒ ∠BOD = ∠BAC

Hence Proved.

Answer 2

Answer:

In triangle OBD and triangle OCD,

OB = OC. { radii of same circle }

OD = OD. { common }

angleODB = angleODC. { each angle = 90°, as OD perpendicular to BC }

triangle OBD congruent to triangle OCD { by RHS rule of congruency }

angle BOD = angle COD. { c.p.c.t }

angle BOC = 2 angle BOD

Now, angle BOC = 2 angle BAC

{ angle form at the center by an arc is double to the angle form on segment }

2 angle BOD = 2 angle BAC

angle BOD = angle BAC, as required.