Advertisements
Advertisements
प्रश्न
O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.
Advertisements
उत्तर
Given: O is the circumcenter of the triangle ABC and D is the midpoint of BC.
To prove: ∠BOD = ∠A
Join OB and OC.
In ΔOBD and ΔCD,
OD = OD ...(Common side)
DB = Dc ...(D is the midpoint of BC)
OB = OC ...(Both are radius of the circle)
By SSS congruence rule, ΔOBD ≅ ΔOCD.
∴ ∠BOD = ∠COD = x (say) ...(By CPCT)
Since, angle subtended by an arc at the center of the circle is twice the angle subtended by it at any other point in the remaining part of the circle, we have:
2∠BAC = ∠BOC
⇒ 2∠BAC = ∠BOD + ∠DOC
⇒ 2∠BAC = x + x
⇒ 2∠BAC = 2x
⇒ ∠BAC = x
⇒ ∠BAC = ∠BOD
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle, is bisected at the point of contact.
Fill in the blanks:
The longest chord of a circle is a __________ of the circle.
In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then find the measure of ∠OAB.
In a cyclic quadrilateral ABCD, if m ∠A = 3 (m ∠C). Find m ∠A.
In Fig., chords AB and CD of the circle intersect at O. AO = 5 cm, BO = 3 cm and CO = 2.5 cm. Determine the length of DO.
In the adjoining figure ‘O’ is the center of the circle, ∠CAO = 25° and ∠CBO = 35°. What is the value of ∠AOB?
Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.
The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA see figure
Draw two acute angles and one obtuse angle without using a protractor. Estimate the measures of the angles. Measure them with the help of a protractor and see how much accurate is your estimate.
In the adjoining figure, AC is a diameter of the circle. AP = 3 cm and PB = 4 cm and QP ⊥ AB. If the area of ΔAPQ is 18 cm2, then the area of shaded portion QPBC is ______.
