NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5

Core Theorems Explained

To solve problems in circles ex 10 5 class 9 ncert, you need to understand three fundamental theorems about circles and cyclic quadrilaterals:

1. The Angle Subtended by an Arc Theorem: The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle. If the central angle is $\angle AOB$ and the angle on the circle is $\angle ACB$, then $\angle AOB = 2\angle ACB$.

1. Angles in the Same Segment Theorem: Angles subtended by the same arc (or chord) in the same segment of a circle are equal. This means if you have points $C$ and $D$ on the major arc, then $\angle ACB = \angle ADB$.

1. Cyclic Quadrilateral Theorem: A quadrilateral is called cyclic if all its four vertices lie on a circle. The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$. Conversely, if the sum of opposite angles of a quadrilateral is $180^\circ$, the quadrilateral is cyclic.

Step-by-Step Guide to Proving a Quadrilateral is Cyclic

1. Identify Key Angles — Locate the opposite pairs of angles in the quadrilateral (e.g., $\angle A$ and $\angle C$, or $\angle B$ and $\angle D$).
2. Apply Geometric Properties — Use known triangle laws (Angle Sum Property, Exterior Angle Theorem) or parallel line properties to express these angles in terms of known variables.
3. Calculate the Angle Sum — Add the opposite angles together and algebraically simplify the expression to show that their sum equals $180^\circ$.
4. Conclude Using Theorem — State the converse of the cyclic quadrilateral theorem: 'Since the sum of a pair of opposite angles is $180^\circ$, the quadrilateral is cyclic.'

Worked Solutions for Key Problem Types

• Example 1: In a circle, points $A, B,$ and $C$ lie on the boundary with centre $O$ such that $\angle BOC = 30^\circ$ and $\angle AOB = 60^\circ$. If $D$ is a point on the circle other than the arc $ABC$, find $\angle ADC$. Step 1: Find the total angle subtended by arc $ABC$ at the centre $O$. $\angle AOC = \angle AOB + \angle BOC = 60^\circ + 30^\circ = 90^\circ$. Step 2: Apply the central angle theorem, which states that the angle subtended by an arc at the centre is double the angle subtended on the remaining part of the circle. Therefore, $\angle AOC = 2 \times \angle ADC$. Step 3: Solve for $\angle ADC$. $\angle ADC = \frac{\angle AOC}{2} = \frac{90^\circ}{2} = 45^\circ$. Final Answer: $\angle ADC = 45^\circ$.
• Example 2: A cyclic quadrilateral $ABCD$ has diagonals intersecting at $E$. If $\angle DBC = 70^\circ$ and $\angle BAC = 30^\circ$, find $\angle BCD$. Step 1: Identify equal angles in the same segment. The arc $BC$ subtends $\angle BAC$ and $\angle BDC$. Therefore, $\angle BDC = \angle BAC = 30^\circ$. Step 2: Look at triangle $BCD$. In $\triangle BCD$, the sum of angles is $180^\circ$. $\angle BCD + \angle BDC + \angle DBC = 180^\circ$. Step 3: Substitute the known values. $\angle BCD + 30^\circ + 70^\circ = 180^\circ \implies \angle BCD + 100^\circ = 180^\circ$. Step 4: Solve for $\angle BCD$. $\angle BCD = 180^\circ - 100^\circ = 80^\circ$. Final Answer: $\angle BCD = 80^\circ$.

Board Exam Tips & Common Pitfalls

Common Mistake: Students often misidentify angles in the same segment. Remember, the two angles must stand on the exact same chord/arc and their vertices must touch the boundary of the same circle segment. Do not assume angles are equal if one vertex is not on the boundary.

Board Presentation Tip: In the CBSE Class 9 exam, write the complete statement of the theorem you are using as a reference in brackets. Writing simply 'by theorem' will lead to a deduction of marks. Always clearly state: 'Angles in the same segment of a circle are equal' or 'Opposite angles of a cyclic quadrilateral sum up to 180 degrees.'

Practice Questions with Solutions

• Q: If the diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. A: Step 1: Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ are diameters intersecting at centre $O$. Step 2: Since $AC$ is a diameter, the angle subtended by it in a semi-circle is $90^\circ$. Thus, $\angle ADC = 90^\circ$ and $\angle ABC = 90^\circ$. Step 3: Similarly, since $BD$ is a diameter, the angle subtended by it in a semi-circle is $90^\circ$. Thus, $\angle DAB = 90^\circ$ and $\angle BCD = 90^\circ$. Step 4: A quadrilateral with all internal angles equal to $90^\circ$ is a rectangle. Final answer: Hence, $ABCD$ is a rectangle.
• Q: If the non-parallel sides of a trapezium are equal, prove that it is cyclic. A: Step 1: Let $ABCD$ be a trapezium with $AB \parallel CD$ and non-parallel sides $AD = BC$. Step 2: Draw perpendiculars $AM \perp CD$ and $BN \perp CD$. Step 3: Compare triangles $\triangle AMD$ and $\triangle BNC$: - $AD = BC$ (Given) - $\angle AMD = \angle BNC = 90^\circ$ (By construction) - $AM = BN$ (Distance between parallel lines $AB$ and $CD$ is constant) Therefore, $\triangle AMD \cong \triangle BNC$ by RHS congruency. Step 4: This gives $\angle ADC = \angle BCD$ (CPCT). Since $AB \parallel CD$, the consecutive interior angles sum to $180^\circ$, i.e., $\angle BAD + \angle ADC = 180^\circ$. Substitute $\angle ADC = \angle BCD$ into the equation: $\angle BAD + \angle BCD = 180^\circ$. Final answer: Since opposite angles of the trapezium sum to $180^\circ$, $ABCD$ is cyclic.
• Q: 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 $\angle ACP = \angle QCD$. A: Step 1: For the left circle, consider chord $AP$. Angles in the same segment are equal, so $\angle ACP = \angle ABP$. Step 2: For the right circle, consider chord $DQ$. Angles in the same segment are equal, so $\angle QCD = \angle QBD$. Step 3: Now look at the intersecting lines $AD$ and $PQ$ which cross at point $B$. The vertically opposite angles must be equal: $\angle ABP = \angle QBD$. Final answer: Since $\angle ACP = \angle ABP$, $\angle QCD = \angle QBD$, and $\angle ABP = \angle QBD$, we conclude that $\angle ACP = \angle QCD$.
• Q: In a circle with four boundary points $A, B, C, D$, the diagonals $AC$ and $BD$ intersect at point $E$ such that $\angle BEC = 130^\circ$ and $\angle ECD = 20^\circ$. Find $\angle BAC$. A: Step 1: Identify that $\angle BEC$ and $\angle DEC$ lie on a straight line $BD$. Therefore, they form a linear pair. $\angle DEC = 180^\circ - \angle BEC = 180^\circ - 130^\circ = 50^\circ$. Step 2: In $\triangle DEC$, the sum of angles is $180^\circ$. $\angle EDC + \angle DEC + \angle ECD = 180^\circ$. $\angle EDC + 50^\circ + 20^\circ = 180^\circ \implies \angle EDC = 180^\circ - 70^\circ = 110^\circ$. Step 3: $\angle BDC$ is the same as $\angle EDC$, which is $110^\circ$. Angles in the same segment of a circle are equal. Therefore, the angle subtended by arc $BC$ are $\angle BAC$ and $\angle BDC$. So, $\angle BAC = \angle BDC = 110^\circ$. Final answer: $\angle BAC = 110^\circ$.

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