Q9 of 22 Page 126

S is the centre of the circumcircle of ΔABC and if AD ⊥ BC, let us prove that ∠BAD = ∠SAC.


Since AD is perpendicular to BC, ∠ADB = 90°


⇒ ∠ABD + ∠BAD = 90 ----------- (1)


By the theorem:-


The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.


------------ (2)


From equation (1) and (2), we get,


------------- (3)


In ΔASC, as AS = SC and we know that angles opposite to equal sides are equal.


⇒ ∠SAC = ∠SCA


Also, ∠SAC + ∠SCA + ∠ASC = 180°


⇒ 2∠SAC + ∠ASC = 180°


⇒ ∠ASC = 180° - 2∠SAC ------------ (4)


From equation (3) and (4), we get,



⇒ ∠BAD = ∠SAC


Hence, proved.


More from this chapter

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7

If the circumcentre of triangle ABC is O; let us prove that ∠OBC + ∠BAC = 90°.

 8

Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, let us prove that ΔABCD is an equilateral triangle.

 10

Two chords AB and CD of a circle with centre O intersect each other at the points P, let us prove that ∠AOD + ∠BOC = 2∠BPC.

If AOD and BOC are supplementary to each other, let us prove that the two chords are perpendicular to each other.

 11

If two chords AB and CD of a circle with centre O, when produced intersect each other at the point P, let us prove that ∠AOC - ∠BOD = 2 ∠BPC.