Q10 of 22 Page 126

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.


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.



Similarly,


∠BOC = 2∠BAC


Adding above equations, we get,


⇒ ∠AOD+∠BOC = 2(∠BAC + ∠DCA) ----------- (1)


In ΔAPC, ∠PAC + ∠PCA = ∠BPC by exterior angles property.


⇒ ∠BAC + ∠DCA = ∠BPC


Hence, proved.


If AOD and BOC are supplementary to each other,


⇒ ∠BAC + ∠DCA = 90


And from above theorem, ∠BPC = 90°


More from this chapter

All 22 →
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.

 9

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

 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.

 12

We drew a circle with the point A of quadrilateral ABCD as centre which passes through the points B,C and D. Let us prove that ∠CBD + ∠ CDB =