#### Chapter 16 Circles R.D. Sharma Solutions for Class 9th Exercise 16.5

Exercise 16.5

1. In Fig. 16.176, Î”ABC is an equilateral triangle. Find m∠BEC.
Solution

2. In Fig. 16.177, Î”PQR is an isosceles triangle with PQ = PR and m ∠PQR = 35°. Find m ∠QSR and m ∠QTR.
Solution

3. In Fig. 16.178, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.
Solution

4. In Fig. 16.179 ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
Solution

5. If ABCD is a cyclic quadrilateral in which AD || BC (Fig. 16.180). Prove that ∠B = ∠C.

Solution

6. In Fig. 16.181, O is the centre of the circle. Find ∠CBD.
Solution

7. In Fig. 16.182, AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.

Solution

8. On a semi-circle with AB as diameter, a point C is taken, so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).

Solution

9. In a cyclic quadrilateral ABCD if AB || CD and ∠B = 70°, find the remaining angles.

Solution

10. In a cyclic quadrilateral ABCD, if m ∠A = 3 (m ∠C). Find m ∠A.

Solution

11. In Fig. 16.183, O is the centre of the circle and ∠DAB = 50° . Calculate the values of x and y.

Solution

12. In Fig. 16.184, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.
Solution

13. In Fig. 16.185, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
Solution

14. In Fig. 16.186, O is the centre of the circle. If ∠CEA = 30°, Find the values of x, y and z.

Solution

15. In Fig. 16.187, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
Solution

16. In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°

Solution

17. In Fig. 16.188, ABCD is a cyclic quadrilateral. Find the value of x.
Solution

18. ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
(iii) ∠BCD = 100° and ∠ABD = 70° find ∠ADB.

Solution

19. Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.
Proof:
We know that the perpendicular bisector of every chord of a circle always passes through the centre.
Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.
Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.

20. Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Solution

21. Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.

Solution

22. If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.

Solution

∠ACB = ∠ADB  ---(2) (Angle in the same segment are equal)
In △ACD and △BDC,
CD = CD (common)
Hence, △ACD ≅ △BDC (SAS congruency criterion)
AC = BD (cpct)
Hence, Proved

23. ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:
(ii) EB = EC.

Solution

24. Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).

Solution

25. Prove that the angle in a segment greater than a semi-circle is less than a right angle.

Solution

26. ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Solution

27. Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half the hypotenuse.

Solution

28. In Fig. 16.189, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution