Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
34 Circle Theorems & Cyclic Quadrilaterals
(A) Prove that the sum of two opposite angles of a cyclic quadrilateral is \(180^\circ\). (3 Marks)
(B) In the given figure, find the value of \(x+y\). (2 Marks)
(B) In the given figure, find the value of \(x+y\). (2 Marks)
Concept:
1
Central Angle Theorem: The angle subtended by an arc at the center is double the angle subtended by it at any remaining part of the circle.
2
Cyclic Quadrilateral: A quadrilateral whose four vertices lie on a circle. Because the central angles making up the whole circle add up to \(360^\circ\), their corresponding inscribed (circumference) angles must add up to half of that (\(180^\circ\)).
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Hint:
1
For the proof (A), draw a circle with centre O and a cyclic quad ABCD. Draw radii OB and OD. Use the central angle theorem on arc BCD and arc DAB. Sum the central angles to 360 to prove the opposite angles sum to 180.
2
For (B), just apply the theorem you proved! Angle A is opposite to Angle C (\(x\)), and Angle B is opposite to Angle D (\(y\)). Find \(x\) and \(y\), then add them.
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Mistakes to Avoid:
!
Assuming Adjacent Angles: Do not sum \(\angle A + \angle B = 180^\circ\). That property belongs to interior angles of parallel lines (like in a parallelogram), not cyclic quadrilaterals! Only opposite angles are supplementary here.
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PART (A) PROOF:
PART (B) SOLUTION:
Given: A cyclic quadrilateral ABCD inscribed in a circle with centre O.
To Prove: \( \angle A + \angle C = 180^\circ \) and \( \angle B + \angle D = 180^\circ \).
Construction: Join OB and OD. [1 M]
Proof:
Consider the arc BCD. It subtends \( \angle BOD \) at the centre and \( \angle BAD \) (or \( \angle A \)) at the circumference.
Therefore, \( \angle BOD = 2\angle A \) ---- (1) [½ M]
Now consider the major arc DAB. It subtends reflex \( \angle BOD \) at the centre and \( \angle BCD \) (or \( \angle C \)) at the circumference.
Therefore, reflex \( \angle BOD = 2\angle C \) ---- (2) [½ M]
Adding (1) and (2):
\( \angle BOD + \text{reflex } \angle BOD = 2\angle A + 2\angle C \)
The angles at the centre of a circle add up to \( 360^\circ \).
\( 360^\circ = 2(\angle A + \angle C) \)
\( \frac{360^\circ}{2} = \angle A + \angle C \)
\( \angle A + \angle C = 180^\circ \) [1 M]
Similarly, by joining OA and OC, we can prove \( \angle B + \angle D = 180^\circ \).
Hence Proved.
PART (B) SOLUTION:
From the figure, ABCD is a cyclic quadrilateral.
Using the property proved above, opposite angles sum to \( 180^\circ \).
\( \angle A + \angle C = 180^\circ \)
\( 60^\circ + x = 180^\circ \implies x = 120^\circ \) [1 M]
Similarly for the other pair:
\( \angle B + \angle D = 180^\circ \)
\( 70^\circ + y = 180^\circ \implies y = 110^\circ \) [½ M]
We need to find the value of \( x + y \):
\( x + y = 120^\circ + 110^\circ = 230^\circ \) [½ M]
Value of (x + y) = 230°
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