Exercise 9.3

1. In the figure A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Instructions

Given that: ∠AOC = ∠AOB + ∠

∠AOC = ° + ° = °

It is known that an angle which is subtended by an arc at the centre of the circle is the angle subtended by that arc at any point on the remaining part of the circle.

Thus, ∠ADC = ∠AOC = × ° = °

So, ∠ADC = 45°

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Instructions

The chord AB is equal to the radius of the circle while OA and OB are the two radii of the circle.

Thus, AB = OA =

Now, considering ΔOAB: it can be said that ΔOAB has sides and thus, it is an triangle.

So, ∠AOC = ° and ∠ACB = ∠AOB. So, ∠ACB = 12 × ° = °

Now, ACBD is a quadrilateral i.e. ∠ADB + ∠ACB = ° (since they are the opposite angles of a cyclic quadrilateral)

Thus, ∠ADB = °- ° = °

So, the angles subtended by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30°, respectively.

3. In the given figure, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

Instructions

The angle which is subtended by an arc at the centre of the circle is the angle subtended by that arc at any point on the remaining part of the circle.

So, the reflex ∠POR = × ∠PQR (we know that ∠PQR = ° )

So, reflex ∠POR = 2 × 100° = ° which further gives us: ∠POR = ° - ° = °

Considering ΔOPR, we get: OP = (they are radii of the circle). Also, ∠OPR = ∠.

We know the sum of the angles in a triangle is equal to 180° i.e. ∠POR + ∠OPR + ∠ = °

∠OPR + ∠OPR = ° - ° (as ∠OPR = ∠)

∠OPR = ° i.e. ∠OPR = °

Thus, ∠OPR = 10°.

4. In the given figure, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.

Instructions

We know that angles in the segment of the circle are , i.e. ∠BAC = ∠

In the ΔABC, the sum of all the interior angles will be 180°. So, ∠ABC + ∠BAC + ∠ACB = °

Putting the values, ∠BAC = 180°- °- ° i.e. ∠BAC = °

Since, ∠BAC = ∠BDC , we get: ∠BDC = °

Thus, ∠BDC = 80°.

5. In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.

Instructions

We know that the angles in the segment of the circle are equal i.e. ∠ BAC = ∠

In ΔCDE, using the exterior angles property of the triangle: ∠ CEB = ∠ CDE + ∠

We have: ∠ DCE = ° which gives us ∠ CDE = °

We can see that from the figure: ∠ CDE = ∠

Thus, ∠ BAC = °

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Instructions

Considering the chord CD: ∠ CBD = ∠ (angles in the are equal)

Thus, ∠ CAD = °

We see that: ∠ BAD will be equal to the sum of angles BAC and .

Thus, ∠ BAD = ∠ BAC + ∠ CAD = ° + ° i.e. ∠ BAD = °

We know that the opposite angles of a cyclic quadrilateral sum up to °.

So, ∠ BCD + ∠ BAD = ° ⇒ ∠ BCD = °

Considering ΔABC: we know AB = i.e. ∠ BCA = ∠ (Angles opposite to sides of a triangle)

Upon solving, we get: ∠ ACD = ° and ∠ ECD = °

Thus, ∠ ACD = 50° and ∠ ECD = 50°.

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Instructions

Drawing a cyclic quadrilateral ABCD inside a circle with centre O such that its diagonal AC and BD are two diameters of the circle.

We know that the angles in the semi-circle are with a value of °

Thus, ∠ ABC = ∠ = ∠ = ∠ DAB = °

So since each internal angle is °, it can be said that the quadrilateral ABCD is a .

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Instructions

If the sum of a pair of angles of a quadrilateral is °, the quadrilateral is cyclic.

Drawing a trapezium ABCD with AB || CD where AD and BC are the non-parallel sides that are equal. Also draw AM ⊥ CD and BN ⊥ CD.

Consider ΔAMD and ΔBNC : AD = (given) and ∠AMD = ∠ = ° while AM = ( distance between two parallel lines is same)

By congruence rule: ΔAMD ≅ Δ. By CPCT: ∠ADC = ∠ (1)

∠BAD and ∠ADC are on the same side of transversal AD i.e. ∠BAD + ∠ADC = ° which can be further written as ∠BAD + ∠ = ° (from (1))

This equation proves that the sum of opposite angles is supplementary. Hence, ABCD is a cyclic quadrilateral.

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 (see the given figure). Prove that ∠ACP = ∠ QCD.

Instructions

Join the chords AP and DQ. For chord AP, we know that angles in the segment are .

So, ∠ PBA = ∠ (i). Similarly, for chord DQ: ∠ DBQ = ∠ (ii).

ABD and PBQ are two line segments which intersect at B, where the vertically opposite angles will be .

Thus, ∠ PBA = ∠ (iii)

From equation (i), equation (ii) and equation (iii) we get: ∠ ACP = ∠

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.

Instructions

Drawing a triangle ABC and then two circles having diameters AB and AC, we have to prove now that D lies on BC and BDC is a straight line.

We know that angles in the semi-circle are i.e. ∠ ADB = ∠ ADC = °

Hence, ∠ ADB + ∠ ADC = °

This implies that ∠ BDC is a angle.

So, D lies on the line BC.

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

Instructions

We know that AC is the common hypotenuse and ∠ B = ∠ D = 90°. Now, it has to be proven that ∠ CAD = ∠ CBD.

Since ∠ ABC and ∠ ADC are °, it can be said that they lie in the .

So, triangles ABC and ADC are in the semi-circle, and the points A, B, C and D are .

Thus, CD is the chord of the circle with centre O.

We know that the angles which are in the same segment of the circle are i.e. ∠ CAD = ∠ .

12. Prove that a cyclic parallelogram is a rectangle.

Instructions

Let ABCD be the cyclic parallelogram. We know that opposite angles of a parallelogram are .

So, ∠A = ∠ and ∠B = ∠ (1)

We know that the sum of either pair of opposite angles of a cyclic quadrilateral is ° ⇒ ∠A + ∠ = °

From equation (1): ∠A + ∠ = 180° ⇒ ∠A = 180° ⇒ ∠A = °

If one of the interior angles of a parallelogram is 90°, all the other angles will also be equal to °.

Since all the angles in the parallelogram are °, we can say that parallelogram ABCD is a .

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Exercise 9.3