- In Fig. 10.36, 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.
Since ABC is an arc of the circle with centre O, which makes an angle AOC = angle AOB + angle BOC
i.e., ∠AOC = ∠AOB + ∠BO = 60° + 30° = 90°
We know ∠AOC = 2 ∠ADC [angle at the centre of a circle is twice the angle at the circumference]
Therefore ∠ADC = 1/2 (∠AOC ) = 1/2 x 90° = 45°
- 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.
Since OA = OB = AB
Therefore, ∆AOB is an equilateral triangle.
⇒∠AOB = 60°
We know that the angle at the centre of a circle is twice the angle at the circumference.
Therefore, ∠AOB = 2 ∠ACB
⇒∠ACB = 1/2 ∠AOB = 1/2 x 60° = 30°
Also, we have , ∠ADB = 1/2reflex ∠AOB = 1/2(360° – 60°) = 1/2 (300°) = 150°
Hence, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc is 30 °
- In Fig. 10.37, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.
Since, angle at the centre of a circle is twice the angle at the circumference.
Therefore, ∠POR = 2 ∠PQR
⇒reflex ∠POR = 2 x 100 ° = 200 °
⇒ ∠POR = 360 ° – 200 ° = 160 °
In ∆OPR, OP = OR [Radii of the same circle]
⇒ ∠OPR = ∠ORP [Angles opposite to equal sides are equal]
and ∠POR = 160 ° [which is proved]
In ∆OPR ,
∠OPR + ∠ORP + ∠PQR = 180 ° [Sum of interior angles of a triangle are equal to 180 °]
⇒160 ° + 2 ∠OPR = 180 °
⇒ 2 ∠OPR = 180 ° – 160 ° = 20 °
⇒ ∠OPR = 10 °
- In Fig. 10.38, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.
∠BAC + ∠ABC + ∠BCA = 180 °
⇒ ∠BAC + 69 ° + 31 ° = 180 °
⇒ 180 ° – 100 ° = 80 °
Since the angles in the same segment are equal.
Therefore, ∠BDC = ∠BAC = 80 °
- In Fig. 10.39, 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.
Given 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°
We have find ∠BAC
Since, ∠CED + ∠CEB = ∠ BED = 180 ° [linear pair]
⇒∠CED + 130 ° = 180 °
⇒∠CED = 180 ° – 130 ° = 50 °
In ∆ECD, we have,
∠EDC + ∠EDC + ∠CED = 180 °
⇒ ∠EDC + 50 ° + 20 ° = 180 °
⇒∠EDC = 180 ° – 50 ° – 20 ° = 110 °
⇒ ∠BDC = ∠EDC = 110 ° [angles in the same segment are equal]
Therefore, ∠BAC = ∠BDC = 110 °
- 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.
Given, ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°.
∠CAD = ∠DBC= 70° [Angles in the same segment]
Therefore, ∠DAB = ∠CAD + ∠BAC = 70° + 30° = 100°
But, ∠DAB + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral]
So, ∠BCD = 180° – 100° = 80°
Now, we have AB = BC
Therefore, ∠BCA = 30° [Opposite angles of an isosceles triangle]
Again, ∠DAB + ∠BCD = 180° [Opposite angles of a cyclic quadrilateral]
⇒ 100° + ∠BCA + ∠ECD = 180° [∵∠BCD = ∠BCA + ∠ECD]
⇒ 100° + 30° + ∠ECD = 180°
⇒ 130° + ∠ECD = 180°
⇒ ∠ECD = 180° – 130° = 50°
Hence, ∠BCD = 80° and ∠ECD = 50°
- If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quadrilateral ABCD.
We have to prove quadrilateral ABCD is a rectangle.
We know, all the radius of the same circle are equal.
Therefore, OA = OB = OC = OD
⇒OA = OC = 1/2 AC
⇒OB = OC = 1/2 BD
⇒AC = BD
Thus, the diagonals of the quadrilateral ABCD are equal and bisect each other.
Quadrilateral ABCD is rectangle.
- If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given : A trapezium ABCD in which AB || CD and AD = BC.
To Prove : ABCD is a cyclic trapezium.
Construction : Draw DE ⊥ AB and CF ⊥ AB.
Proof : In ∆DEA and ∆CFB, we have
AD = BC [Given]
∠DEA = ∠CFB = 90° [DE ⊥ AB and CF ⊥ AB]
DE = CF [Distance between parallel lines remains constant]
∴ ∆DEA ≅ ∆CFB [RHS axiom]
⇒ ∠A = ∠B …(i) [CPCT]
and, ∠ADE = ∠BCF ..(ii) [CPCT]
Since, ∠ADE = ∠BCF [From (ii)]
⇒ ∠ADE + 90° = ∠BCF + 90°
⇒ ∠ADE + ∠CDE = ∠BCF + ∠DCF
⇒ ∠D = ∠C ……………………………….(iii)
[∠ADE + ∠CDE = ∠D, ∠BCF + ∠DCF = ∠C]
∴ ∠A = ∠B and ∠C = ∠D [From (i) and (iii)] (iv)
∠A + ∠B + ∠C + ∠D = 360° [Sum of the angles of a quadrilateral is 360°]
⇒ 2(∠B + ∠D) = 360° [Using (iv)]
⇒ ∠B + ∠D = 180°
⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°.
⇒ ABCD is a cyclic trapezium
- 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 Fig. 10.40). Prove that ∠ ACP = ∠ QCD.
Since angles in the same segment are equal.
Therefore, ∠ACP = ∠ABP ……………….(1)
and ∠QCD = ∠QBD ……………………(2)
Also ∠ABP = ∠QBD ……………………..(3) [vertically opposite angles]
Therefore, from (1), (2) and (3), we have
∠ACP = ∠QCD
- 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.
Given: Two circles are drawn with sides AB and AC of ∆ABC as diameters . The circle intersect at D
To prove: D lies on BC
Join A and D
Proof: since AB and AC are the diameters of the two circles[given]
Therefore, ∠ADB = 90° [angles are in a semi circle]
and ∠ADC = 90° [angles in a semi circle]
On adding, we get,
⇒∠ADB + ∠ADC = 90 + 90 = 180
⇒BDC is a straight line
Hence D lies on BC
- ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠ CBD.
∆ABC and ∆ADC are right angles with common hypotenuse AC. Draw a circle AC as diameter passing through B and D . Join BD
Clearly, ∠CAD = ∠CBD [since angles in the same segment are equal]
- Prove that a cyclic parallelogram is a rectangle.
Given: ABCD is a parallelogram inscribed in a circle
To prove: ABCD is a rectangle
Proof: Since ABCD is a cyclic parallelogram
Therefore, ∠A + ∠C = 180° ………………..(1)
But ∠A = ∠C ………………..(2)
From (1) and (2), we have
∠A = ∠C = 90°
Similarly, ∠B = ∠D = 90°
Therefore, each angle of ABCD is of 90°
Hence, ABCD is a rectangle.