O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A

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Solution

We have to prove that

Since, circumcenter is the intersection of perpendicular bisectors of each side of the triangle.

Now according to figure A, B, C are the vertices of ΔABC

In , is perpendicular bisector of BC

So, BD = CD
OB = OC (Radius of the same circle)
And,

OD = OD (Common)

Therefore,
$△ B D O ≅ △ C D O (SSS congruency criterion)$

$∠ B O D = ∠ C O D (by cpct)$

We know that angle formed any chord of the circle at the center is twice of the angle formed at the circumference by same chord

Therefore,

Therefore,

Hence proved

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Similar questions

In the figure, O is the circumcentre of the triangle ABC, AB = AC and the line OD is perpendicular to the side BC. If BC = 24 cm and OD = 5 cm, then find the circumradius.

Q. Question 7
O is the circumcentre of the

Δ A B C

and D is the mid-point of the base BC. Prove that

∠ B O D = ∠ A

O

is the centroid of

△ A B C

and

D

is mid point of base

B C

then prove

∠ B O D = ∠ A .

Q. Fill In The Blanks

If O is the circumcentre of ΔABC and D is the mid-point of the base BC, then ∠BOD = _______________.

Q. In an acute-angled triangle

A B C,

a point

D

lies on the segment

B C .

Let

O_{1}, O_{2}

denote the circumcentres of

Δ A B D

and

Δ A C D,

respectively. Prove that the line joining the circumcentre of

Δ A B C

and the orthocentre of

Δ O_{1} O_{2} D

is parallel to

B C .

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