NCERT Solution Class 9th Maths Chapter – 9 Circles Examples

Solution – Given that AB and CD are two chords of a circle, with centre O intersecting at a point E. PQ is a diameter through E, such that ∠ AEQ = ∠ DEQ (see Fig.9.11). You have to prove that AB = CD. Draw perpendiculars OL and OM on chords AB and CD, respectively. Now
∠ LOE = 180° – 90° – ∠ LEO = 90° – ∠ LEO
(Angle sum property of a triangle)
= 90° – ∠ AEQ = 90° – ∠ DEQ
= 90° – ∠ MEO = ∠ MOE

Solution – Join OC, OD and BC.
Triangle ODC is equilateral (Why?)
Therefore, ∠ COD = 60°
Now, ∠ CBD = 1/2 ∠ COD ( Theorem 9.7)
This gives ∠ CBD = 30°
Again, ∠ ACB = 90° (Why ?)
So, ∠ BCE = 180° – ∠ ACB = 90°
Which gives ∠ CEB = 90° – 30° = 60°, i.e., ∠ AEB = 60°

Solution – ∠ CAD = ∠ DBC = 55°
(Angles in the same segment)
Therefore, ∠ DAB = ∠ CAD + ∠ BAC
= 55° + 45° = 100°
But ∠ DAB + ∠ BCD = 180°
(Opposite angles of a cyclic quadrilateral)
So, ∠ BCD = 180° – 100° = 80°

Solution – Join AB.
∠ ABD = 90° (Angle in a semicircle)
∠ ABC = 90° (Angle in a semicircle)
So, ∠ ABD + ∠ ABC = 90° + 90° = 180°
Therefore, DBC is a line. That is B lies on the line segment DC

Solution – In Fig. 9.22, ABCD is a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles A, B, C and D respectively form a quadrilateral EFGH.
Now, ∠ FEH = ∠ AEB = 180° – ∠ EAB – ∠ EBA (Why ?)
= 180° –1/2 (∠ A + ∠ B)
and ∠ FGH = ∠ CGD = 180° – ∠ GCD – ∠ GDC (Why ?)
= 180° –1/2 (∠ C + ∠ D)

Therefore, ∠ FEH + ∠ FGH = 180° –1/2 (∠ A + ∠ B) + 180° –1/2 (∠ C + ∠ D)
= 360° –1/2 (∠ A+ ∠ B +∠ C +∠ D) = 360° – 1/2 × 360°
= 360° – 180° = 180°
Therefore, by Theorem 9.11, the quadrilateral EFGH is cyclic