Q3 of 63 Page 150

O is the circumcircle of triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD = ∠A.


Given in ΔABC, O is the circum circle and D is the mid-point of the base BC.


We have to prove that ∠BOD = ∠A.


Construction:


Join OB and OC


Proof:


Consider ΔOBD and ΔOCD,


⇒ OB = OC [Radii of same circle]


⇒ BD = DC [D is mid-point of BC]


⇒ OD = OD [Common side]


By SSS congruence rule,


ΔOBD ≅ ΔOCD


By CPCT,


⇒ ∠BOD = ∠COD


∴ ∠BOD = 1/2 ∠BOC … (1)


Arc BC subtends ∠BOC at the centre and ∠BAC at point A in the remaining part of the circle.


∴ ∠BAC = 1/2 ∠BOC … (2)


From (1) and (2),


⇒∠BOD = ∠BAC


∴ ∠BOD = ∠A


Hence proved


More from this chapter

All 63 →
1

Write True/False for each and write the reason also for your answer.

(i) The angles subtended at any two points lying of the circle by every chord are equal.


(ii) In figure, AB is a diameter of a circle and C is any point on the circle. Then AC2 + BC2 = AB2.


(iii) In figure 12.48, if ∠ADE = 120°, then ∠EAB = 30°.


(iv) In figure 12.48, ∠CAD = ∠CED.


 2

In figure, ∠ABC = 45°. Then prove that OA ⊥ OC.

 4

On a common hypotenuse AB two right angled triangles ACB and ADB are drawn such that they lie on the opposite sides. Prove that ∠BAC = ∠BDC.

 5

Two chords AB and AC of a circle subtend angles 90° and 150° respectively on its centre. Find ∠BAC if AB and AC lie on opposite side of the centre.