Circle

Exercise 10.5 Part 3

Question 9: Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that ∠ACP = ∠QCD.

Answer: Let us join A to P to make chord AP and then join D to Q to make chord DQ

For chord AP;
∠ACP = ∠PBA ………….(1)
(Angles in the same segment)
Similarly, for chord DQ;
∠DCQ = ∠DBQ …………(2)

We also have;
∠PBA = ∠DBQ (Vertically opposite angles) …………..(3)
From equations (1), (2) and (3);
∠ACP = ∠DCQ proved

Question 10: If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer: ABC is a triangle. AB and AC are diameters of two circles which intersect at A and D.

∠ADC = ∠ADB = 90° (angles in semi-circle)
As ∠ADC + ∠ADB = 180°So, AB is a straight line
This proves that D lies on AB and is the point of intersection for two circles.

Question 11: ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Answer: Given ABC and ADC are right triangles with common hypotenuse AC.
To Prove ∠CAD = ∠CBD

Let us take chord DC;
∠CAD = ∠CBD
(Angles in the same segment).

Question 12: Prove that a cyclic parallelogram is a rectangle.

Answer: Opposite angles of a cyclic parallelogram are supplementary.
If opposite angles are ∠1 and ∠2
Then; ∠1 + ∠2 = 180°
Moreover, opposite angles of a cyclic parallelogram are equal
Hence, ∠1 = ∠2 = 90°
Thus all angles of cyclic parallelogram are right angles
Hence, a cyclic parallelogram is a rectangle proved.