RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5

Other Exercises

Question 1.
In the figure, ∆ABC is an equilateral triangle. Find m ∠BEC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q1.1
Solution:
∵ ∆ABC is an equilateral triangle
∴ A = 60°
∵ ABEC is a cyclic quadrilateral
∴ ∠A + ∠E = 180° (Sum of opposite angles)
⇒ 60° + ∠E = 180°
⇒ ∠E = 180° – 60° = 120°
∴ m ∠BEC = 120°

Question 2.
In the figure, ∆PQR is an isosceles triangle with PQ = PR and m ∠PQR = 35°. Find m ∠QSR and m ∠QTR.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q2.1
Solution:
In the figure, ∆PQR is an isosceles PQ = PR
∠PQR = 35°
∴ ∠PRQ = 35°
But ∠PQR + ∠PRQ + ∠QPR = 180° (Sum of angles of a triangle)
⇒ 35° + 35° + ∠QPR = 180°
⇒ 70° + ∠QPR = 180°
∴ ∠QPR = 180° – 70° = 110°
∵ ∠QSR = ∠QPR (Angle in the same segment of circles)
∴ ∠QSR = 110°
But PQTR is a cyclic quadrilateral
∴ ∠QTR + ∠QPR = 180°
⇒ ∠QTR + 110° = 180°
⇒ ∠QTR = 180° -110° = 70°
Hence ∠QTR = 70°

Question 3.
In the figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q3.1
Solution:
In the figure, O is the centre of the circle ∠BOD =160°
ABCD is the cyclic quadrilateral
∵ Arc BAD subtends ∠BOD is the angle at the centre and ∠BCD is on the other part of the circle
∴ ∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD
⇒ x = \(\frac { 1 }{ 2 }\) x 160° = 80°
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180°
⇒ y + x = 180°
⇒ y + 80° = 180°
⇒ y =180°- 80° = 100°
∴ x = 80°, y = 100°

Question 4.
In the figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q4.1
Solution:
In a circle, ABCD is a cyclic quadrilateral ∠BCD = 100° and ∠ABD = 70°
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180° (Sum of opposite angles)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q4.2
⇒ ∠A + 100°= 180°
∠A = 180°- 100° = 80°
Now in ∆ABD,
∠A + ∠ABD + ∠ADB = 180°
⇒ 80° + 70° + ∠ADB = 180°
⇒ 150° +∠ADB = 180°
∴ ∠ADB = 180°- 150° = 30°
Hence ∠ADB = 30°

Question 5.
If ABCD is a cyclic quadrilateral in which AD || BC. Prove that ∠B = ∠C.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q5.1
Solution:
Given : ABCD is a cyclic quadrilateral in which AD || BC
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q5.2
To prove : ∠B = ∠C
Proof : ∵ AD || BC
∴ ∠A + ∠B = 180°
(Sum of cointerior angles)
But ∠A + ∠C = 180°
(Opposite angles of the cyclic quadrilateral)
∴ ∠A + ∠B = ∠A + ∠C
⇒ ∠B = ∠C
Hence ∠B = ∠C

Question 6.
In the figure, O is the centre of the circle. Find ∠CBD.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q6.1
Solution:
Arc AC subtends ∠AOC at the centre and ∠APC at the remaining part of the circle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q6.2
∴ ∠APC = \(\frac { 1 }{ 2 }\) ∠AOC
= \(\frac { 1 }{ 2 }\) x 100° = 50°
∵ APCB is a.cyclic quadrilateral,
∴ ∠APC + ∠ABC = 180°
⇒ 50° + ∠ABC = 180° ⇒ ∠ABC =180°- 50°
∴ ∠ABC =130°
But ∠ABC + ∠CBD = 180° (Linear pair)
⇒ 130° + ∠CBD = 180°
⇒ ∠CBD = 180°- 130° = 50°
∴ ∠CBD = 50°

Question 7.
In the figure, AB and CD are diameiers of a circle with centre O. If ∠OBD = 50°, find ∠AOC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q7.1
Solution:
Two diameters AB and CD intersect each other at O. AC, CB and BD are joined
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q7.2
∠DBA = 50°
∠DBA and ∠DCA are in the same segment
∴ ∠DBA = ∠DCA = 50°
In ∆OAC, OA = OC (Radii of the circle)
∴ ∠OAC = ∠OCA = ∠DCA = 50°
and ∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)
⇒ 50° + 50° + ∠AOC = 180°
⇒ 100° + ∠AOC = 180°
⇒ ∠AOC = 180° – 100° = 80°
Hence ∠AOC = 80°

Question 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:
A semicircle with AB as diameter
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q8.1
∠ CAB = 30°
∠ACB = 90° (Angle in a semi circle)
But ∠CAB + ∠ACB + ∠ABC = 180°
⇒ 30° + 90° + ∠ABC – 180°
⇒ 120° + ∠ABC = 180°
∴ ∠ABC = 180°- 120° = 60°
Hence m ∠ACB = 90°
and m ∠ABC = 60°

Question 9.
In a cyclic quadrilateral ABCD, if AB || CD and ∠B = 70°, find the remaining angles.
Solution:
In a cyclic quadrilateral ABCD, AB || CD and ∠B = 70°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q9.1
∵ ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180°
⇒ 70° + ∠D = 180°
⇒ ∠D = 180°-70° = 110°
∵ AB || CD
∴ ∠A + ∠D = 180° (Sum of cointerior angles)
∠A+ 110°= 180°
⇒ ∠A= 180°- 110° = 70°
Similarly, ∠B + ∠C = 180°
⇒ 70° + ∠C- 180° ‘
⇒ ∠C = 180°-70°= 110°
∴ ∠A = 70°, ∠C = 110°, ∠D = 110°

Question 10.
In a cyclic quadrilateral ABCD, if m ∠A = 3(m ∠C). Find m ∠A.
Solution:
In cyclic quadrilateral ABCD, m ∠A = 3(m ∠C)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q10.1
∵ ABCD is a cyclic quadrilateral,
∴ ∠A + ∠C = 180°
⇒ 3 ∠C + ∠C = 180° ⇒ 4∠C = 180°
⇒ ∠C = \(\frac { { 180 }^{ \circ } }{ 4 }\)  = 45°
∴ ∠A = 3 x 45°= 135°
Hence m ∠A =135°

Question 11.
In the figure, O is the centre of the circle and ∠DAB = 50°. Calculate the values of x and y.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q11.1
Solution:
In the figure, O is the centre of the circle ∠DAB = 50°
∵ ABCD is a cyclic quadrilateral
∴ ∠A + ∠C = 180°
⇒ 50° + y = 180°
⇒ y = 180° – 50° = 130°
In ∆OAB, OA = OB (Radii of the circle)
∴ ∠A = ∠OBA = 50°
∴ Ext. ∠DOB = ∠A + ∠OBA
x = 50° + 50° = 100°
∴ x= 100°, y= 130°

Question 12.
In the figure, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q12.1
Solution:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q12.2
60° + ∠ABC + 20° = 180°
∠ABC + 80° = 180°
∴ ∠ABC = 180° -80°= 100°
∵ ABCD is a cyclic quadrilateral,
∴ ∠ABC + ∠ADC = 180°
100° + ∠ADC = 180°
∴ ∠ADC = 180°- 100° = 80°

Question 13.
In the figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q13.1
Solution:
In a circle, ∆ABC is an equilateral triangle
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q13.2
∴ ∠A = 60°
∵ ∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 60°
∵ BECD is a cyclic quadrilateral
∴ ∠BDC + ∠BEC = 180°
⇒ 60° + ∠BEC = 180°
⇒ ∠BEC = 180°-60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120°

Question 14.
In the figure, O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q14.1
Solution:
∠AEC and ∠ADC are in the same segment
∴ ∠AEC = ∠ADC = 30°
∴ z = 30°
ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180°
⇒ x + z = 180°
⇒ x + 30° = 180°
⇒ x = 180° – 30° = 150°
Arc AC subtends ∠AOB at the centre and ∠ADC at the remaining part of the circle
∴ ∠AOC = 2∠D = 2 x 30° = 60°
∴ y = 60°
Hence x = 150°, y – 60° and z = 30°

Question 15.
In the figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q15.1
Solution:
In the figure, two circles intersect each other at C and D
∠BAD = 78°, ∠DCF = x, ∠DEF = y
ABCD is a cyclic quadrilateral
∴ Ext. ∠DCF = its interior opposite ∠BAD
⇒ x = 78°
In cyclic quadrilateral CDEF,
∠DCF + ∠DEF = 180°
⇒ 78° + y = 180°
⇒ y = 180° – 78°
y = 102°
Hence x = 78°, and y- 102°

Question 16.
In a cyclic quadrilateral ABCD, if ∠A – ∠C = 60°, prove that the smaller of two is 60°.
Solution:
In cyclic quadrilateral ABCD,
∠A – ∠C = 60°
But ∠A + ∠C = 180° (Sum of opposite angles)
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q16.1
Adding, 2∠A = 240° ⇒ ∠A = \(\frac { { 62 }^{ \circ } }{ 2 }\)  = 120° and subtracting
2∠C = 120° ⇒ ∠C = \(\frac { { 120 }^{ \circ } }{ 2 }\)  = 60°
∴ Smaller angle of the two is 60°.

Question 17.
In the figure, ABCD is a cyclic quadrilateral. Find the value of x.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q17.1
Solution:
∠CDE + ∠CDA = 180° (Linear pair)
⇒ 80° + ∠CDA = 180°
⇒ ∠CDA = 180° – 80° = 100°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q17.2
In cyclic quadrilateral ABCD,
Ext. ∠ABF = Its interior opposite angle ∠CDA = 100°
∴ x = 100°

Question 18.
ABCD is a cyclic quadrilateral in which:
(i) BC || AD, ∠ADC =110° and ∠B AC = 50°. Find ∠DAC.
(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
(iii) ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
Solution:
(i) In the figure,
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q18.1
ABCD is a cyclic quadrilateral and AD || BC, ∠ADC = 110°
∠BAC = 50°
∵ ∠B + ∠D = 180° (Sum of opposite angles)
⇒ ∠B + 110° = 180°
∴ ∠B = 180°- 110° = 70°
Now in ∆ABC,
∠CAB + ∠ABC + ∠BCA = 180° (Sum of angles of a triangle)
⇒ 50° + 70° + ∠BCA = 180°
⇒ 120° + ∠BCA = 180°
⇒ ∠BCA = 180° – 120° = 60°
But ∠DAC = ∠BCA (Alternate angles)
∴ ∠DAC = 60°
(ii) In cyclic quadrilateral ABCD,
Diagonals AC and BD are joined ∠DBC = 80°, ∠BAC = 40°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q18.2
Arc DC subtends ∠DBC and ∠DAC in the same segment
∴ ∠DBC = ∠DAC = 80°
∴ ∠DAB = ∠DAC + ∠CAB = 80° + 40° = 120°
But ∠DAC + ∠BCD = 180° (Sum of opposite angles of a cyclic quad.)
⇒ 120° +∠BCD = 180°
⇒ ∠BCD = 180°- 120° = 60°
(iii) In the figure, ABCD is a cyclic quadrilateral BD is joined
∠BCD = 100°
and ∠ABD = 70°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q18.3
∠A + ∠C = 180° (Sum of opposite angles of cyclic quad.)
∠A+ 100°= 180°
⇒ ∠A= 180°- 100°
∴ ∠A = 80°
Now in ∆ABD,
∠A + ∠ABD + ∠ADB = 180° (Sum of angles of a triangle)
⇒ 80° + 70° + ∠ADB = 180°
⇒ 150° +∠ADB = 180°
⇒ ∠ADB = 180°- 150° = 30°
∴ ∠ADB = 30°

Question 19.
Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.
Solution:
Given : ABCD is a rhombus. Four circles are drawn on the sides AB, BC, CD and DA respectively
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q19.1
To prove : The circles pass through the point of intersection of the diagonals of the rhombus ABCD
Proof: ABCD is a rhombus whose diagonals AC and BD intersect each other at O
∵ The diagonals of a rhombus bisect each other at right angles
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Now when ∠AOB = 90°
and a circle described on AB as diameter will pass through O
Similarly, the circles on BC, CD and DA as diameter, will also pass through O

Question 20.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that is diagonals are equal.
Solution:
Given : In cyclic quadrilateral ABCD, AB = CD
AC and BD are the diagonals
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q20.1
To prove : AC = BC
Proof: ∵ AB = CD
∴ arc AB = arc CD
Adding arc BC to both sides, then arc AB + arc BC = arc BC + arc CD
⇒ arc AC = arc BD
∴ AC = BD
Hence diagonal of the cyclic quadrilateral are equal.

Question 21.
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:
Given : In ∆ABC, circles are drawn on sides AB and AC
To prove : Circles drawn on AB and AC intersect at D which lies on BC, the third side
Construction : Draw AD ⊥ BC
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q21.1
Proof: ∵ AD ⊥ BC
∴ ∠ADB = ∠ADC = 90°
So, the circles drawn on sides AB and AC as diameter will pass through D
Hence circles drawn on two sides of a triangle pass through D, which lies on the third side.

Question 22.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Solution:
In the figure, ABCD is a trapezium in which AD || BC and ∠B = 70°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q22.1
∵ AD || BC
∴ ∠A + ∠B = 180° (Sum of cointerior angles)
⇒ ∠A + 70° = 180°
⇒ ∠A= 180°- 70° = 110°
∴ ∠A = 110°
But ∠A + ∠C = 180° and ∠B + ∠D = 180° (Sum of opposite angles of a cyclic quadrilateral)
∴ 110° + ∠C = 180°
⇒ ∠C = 180°- 110° = 70°
and 70° + ∠D = 180°
⇒ ∠D = 180° – 70° = 110°
∴ ∠A = 110°, ∠C = 70° and ∠D = 110°

Question 23.
In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q23.1
Solution:
In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are drawn ∠DBC = 55° and ∠BAC = 45°
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q23.2
∵ ∠BAC and ∠BDC are in the same segment
∴ ∠BAC = ∠BDC = 45°
Now in ABCD,
∠DBC + ∠BDC + ∠BCD = 180° (Sum of angles of a triangle)
⇒ 55° + 45° + ∠BCD = 180°
⇒ 100° + ∠BCD = 180°
⇒ ∠BCD = 180° – 100° = 80°
Hence ∠BCD = 80°

Question 24.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Solution:
Given : ABCD is a cyclic quadrilateral
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q24.1
To prove : The perpendicular bisectors of the sides are concurrent
Proof : ∵ Each side of the cyclic quadrilateral is a chord of the circle and perpendicular of a chord passes through the centre of the circle
Hence the perpendicular bisectors of each side will pass through the centre O
Hence the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent

Question 25.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Solution:
Given : ABCD is a cyclic rectangle and diagonals AC and BD intersect each other at O
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q25.1
To prove : O is the point of intersection is the centre of the circle.
Proof : Let O be the centre of the circle- circumscribing the rectangle ABCD
Since each angle of a rectangle is a right angle and AC is the chord of the circle
∴ AC will be the diameter of the circle Similarly, we can prove that diagonal BD is also the diameter of the circle
∴ The diameters of the circle pass through the centre
Hence the point of intersection of the diagonals of the rectangle is the centre of the circle.

Question 26.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:
(i) AD || BC
(ii) EB = EC.
Solution:
Given : ABCD is a cyclic quadrilateral in which sides BA and CD are produced to meet at E and EA = ED
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q26.1
To prove :
(i) AD || BC
(ii) EB = EC
Proof: ∵ EA = ED
∴ In ∆EAD
∠EAD = ∠EDA (Angles opposite to equal sides)
In a cyclic quadrilateral ABCD,
Ext. ∠EAD = ∠C
Similarly Ext. ∠EDA = ∠B
∵ ∠EAD = ∠EDA
∴ ∠B = ∠C
Now in ∆EBC,
∵ ∠B = ∠C
∴ EC = EB (Sides opposite to equal sides)
and ∠EAD = ∠B
But these are corresponding angles
∴ AD || BC

Question 27.
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Solution:
Given : A segment ACB shorter than a semicircle and an angle ∠ACB inscribed in it
To prove : ∠ACB < 90°
Construction : Join OA and OB
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q27.1
Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle ∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB But ∠AOB > 180° (Reflex angle)
∴ ∠ACB > \(\frac { 1 }{ 2 }\) x [80°
⇒ ∠ACB > 90°

Question 28.
Prove that the angle in a segment greater than a semi-circle is less than a right angle
Solution:
Given : A segment ACB, greater than a semicircle with centre O and ∠ACB is described in it
To prove : ∠ACB < 90°
Construction : Join OA and OB
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q28.1
Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle
∴ ∠ACB =\(\frac { 1 }{ 2 }\) ∠AOB
But ∠AOB < 180° (A straight angle) 1
∴ ∠ACB < \(\frac { 1 }{ 2 }\) x 180°
⇒ ∠ACB <90°
Hence ∠ACB < 90°

Question 29.
Prove that the line segment joining the mid-point of the hypotenuse of a rijght triangle to its opposite vertex is half of the hypotenuse.
Solution:
Given : In a right angled ∆ABC
∠B = 90°, D is the mid point of hypotenuse AC. DB is joined.
To prove : BD = \(\frac { 1 }{ 2 }\) AC
Construction : Draw a circle with centre D and AC as diameter
RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 Q29.1
Proof: ∵ ∠ABC = 90°
∴ The circle drawn on AC as diameter will pass through B
∴ BD is the radius of the circle
But AC is the diameter of the circle and D is mid point of AC
∴ AD = DC = BD
∴ BD= \(\frac { 1 }{ 2 }\) AC

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