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NCERT Exemplar · Class 9 Mathematics Circles

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EXERCISE 10.4 1–10 (part 5 of 6)

  1. Exercise 1

    If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q1 Let the equal chords \(\displaystyle AB, CD\) meet at \(\displaystyle P\); \(\displaystyle OM \perp AB\), \(\displaystyle ON \perp CD\) (\(\displaystyle O\) = centre). \[AB = CD \quad \text{(given)} \] \[OM = ON \quad \text{(equal chords are equidistant from the centre)} \] In \(\displaystyle \triangle OMP\) and \(\displaystyle \triangle ONP\): \[\angle OMP = \angle ONP = 90^\circ, \quad OP = OP, \quad OM = ON \] \[\triangle OMP \cong \triangle ONP \quad \text{(RHS)} \] \[PM = PN \] \[AM = MB = CN = ND \quad \text{(perpendicular from centre bisects the chord; } AB = CD\text{)} \] \[AP = AM - PM = ND - PN = DP \] \[BP = MB + PM = CN + PN = CP \]Answer: \(\displaystyle AP = DP\) and \(\displaystyle BP = CP\).

  2. Exercise 2

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

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q2 Let \(\displaystyle ABCD\) be the trapezium, \(\displaystyle AB \parallel DC\), \(\displaystyle AD = BC\). Draw \(\displaystyle AM \perp DC\), \(\displaystyle BN \perp DC\) (\(\displaystyle M, N\) on \(\displaystyle DC\)). \[AM = BN \quad \text{(distance between the parallels } AB, DC\text{)} \] In \(\displaystyle \triangle AMD\) and \(\displaystyle \triangle BNC\): \[\angle AMD = \angle BNC = 90^\circ, \quad AD = BC \ \text{(given)}, \quad AM = BN \] \[\triangle AMD \cong \triangle BNC \quad \text{(RHS)} \] \[\angle ADC = \angle BCD \] \[\angle A + \angle D = 180^\circ \quad \text{(co-interior angles, } AB \parallel DC\text{, transversal } AD\text{)} \] \[\angle A + \angle C = 180^\circ \]Answer: \(\displaystyle \angle A + \angle C = 180^\circ\), so \(\displaystyle ABCD\) is cyclic.

  3. Exercise 3

    If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC , prove that P,Q,R\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{R} and D are concyclic.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q3 \[RQ \parallel BC, \ RQ = \tfrac12 BC \quad \text{(midpoint theorem, in } \triangle ABC\text{)} \] \[BP = \tfrac12 BC \quad (P \text{ midpoint of } BC) \] \[RQ = BP, \ RQ \parallel BP \ \Rightarrow \ BPQR \text{ is a parallelogram} \] \[\angle RQP = \angle RBP = \angle B \quad \text{(opposite angles of a parallelogram; } \angle RBP = \angle ABC\text{)} \] \[RD = RB = \tfrac12 AB \quad \text{(} R \text{ is the mid-point of the hypotenuse of right } \triangle ABD, \ \angle ADB = 90^\circ\text{)} \] \[\angle RDB = \angle RBD = \angle B \quad (\triangle RBD \text{ isosceles, } RD = RB) \] \[\angle RDP = 180^\circ - \angle RDB = 180^\circ - \angle B \quad (B, D, P \text{ collinear}) \] \[\angle RDP + \angle RQP = 180^\circ \] Answer: \(\displaystyle \angle RQP + \angle RDP = 180^\circ \Rightarrow P, Q, R, D \text{ are concyclic.} \)

  4. Exercise 4

    ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q . Prove that P,Q,C\displaystyle \mathrm{P}, \mathrm{Q}, \mathrm{C} and D are concyclic.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q4 \[A, B, Q, P \text{ concyclic (given circle)} \] \[\angle APQ + \angle ABQ = 180^\circ \quad \text{(opposite angles of cyclic quad. } ABQP\text{)} \] \[\angle QPD = 180^\circ - \angle APQ = \angle ABQ \quad (A, P, D \text{ collinear}) \] \[\angle ABQ = \angle ABC \quad (Q \text{ on } BC) \] \[\angle ABC + \angle BCD = 180^\circ \quad \text{(co-interior angles, } AB \parallel DC\text{)} \] \[\angle QPD = 180^\circ - \angle BCD = 180^\circ - \angle QCD \quad (Q \text{ on } BC) \] \[\angle QPD + \angle QCD = 180^\circ \] Answer: \(\displaystyle \angle QPD + \angle QCD = 180^\circ \Rightarrow P, Q, C, D \text{ are concyclic.} \)

  5. Exercise 5

    Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q5 Let the bisector of \(\displaystyle \angle A\) meet the circumcircle of \(\displaystyle \triangle ABC\) again at \(\displaystyle M\). \[\angle BAM = \angle CAM \quad (AM \text{ bisects } \angle A) \] \[\text{arc } BM = \text{arc } CM \quad \text{(equal inscribed angles subtend equal arcs)} \] \[BM = CM \quad \text{(equal arcs} \Rightarrow \text{equal chords)} \] So \(\displaystyle M\) is equidistant from \(\displaystyle B\) and \(\displaystyle C\), i.e. \(\displaystyle M\) lies on the perpendicular bisector of \(\displaystyle BC\) — and \(\displaystyle M\) lies on the bisector of \(\displaystyle \angle A\) and on the circumcircle.Answer: The bisector of \(\displaystyle \angle A\) meets the perpendicular bisector of \(\displaystyle BC\) at \(\displaystyle M\), which lies on the circumcircle.

  6. Exercise 6

    If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see Fig.10.18), prove that arc⁡CXA+arc⁡DZB=arc⁡AYD+arc⁡BWC=\displaystyle \operatorname{arc} \mathrm{CXA}+\operatorname{arc} \mathrm{DZB}=\operatorname{arc} \mathrm{AYD}+\operatorname{arc} \mathrm{BWC}= semicircle. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-4_Q6

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q6 Let \(\displaystyle AB, CD\) meet at \(\displaystyle O\), \(\displaystyle \angle AOC = 90^\circ\). In \(\displaystyle \triangle BOC\), \(\displaystyle \angle AOC\) is the exterior angle at \(\displaystyle O\): \[\angle AOC = \angle OBC + \angle OCB \quad \text{(exterior angle of a triangle)} \] \[\angle OBC = \angle ABC = \tfrac12\,\text{arc } AXC, \quad \angle OCB = \angle DCB = \tfrac12\,\text{arc } DZB \quad \text{(inscribed angle} = \tfrac12 \text{arc)} \] \[90^\circ = \tfrac12\left(\text{arc } AXC + \text{arc } DZB\right) \] \[\text{arc } CXA + \text{arc } DZB = 180^\circ \] \[\text{arc } AYD + \text{arc } DZB + \text{arc } BWC + \text{arc } CXA = 360^\circ \quad \text{(whole circle)} \] \[\text{arc } AYD + \text{arc } BWC = 360^\circ - 180^\circ = 180^\circ \]Answer: \(\displaystyle \text{arc } CXA + \text{arc } DZB = \text{arc } AYD + \text{arc } BWC = 180^\circ\) (each a semicircle).

  7. Exercise 7

    If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, prove that PA is angle bisector of ∠BPC\displaystyle \angle \mathrm{BPC}.

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    Since \(\displaystyle AB=BC=CA\), equal chords cut off equal arcs: \[\overset{\frown}{AB}=\overset{\frown}{BC}=\overset{\frown}{CA} \] \[\overset{\frown}{AB}+\overset{\frown}{BC}+\overset{\frown}{CA}=360^\circ \ \Rightarrow\ \overset{\frown}{AB}=\overset{\frown}{BC}=\overset{\frown}{CA}=120^\circ \] Arcs \(\displaystyle AB\) and \(\displaystyle CA\) do not contain \(\displaystyle P\): NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q7 \[\angle APB=\tfrac12\overset{\frown}{AB}=60^\circ \] \[\angle APC=\tfrac12\overset{\frown}{CA}=60^\circ \] \[\Rightarrow \angle APB=\angle APC \] Answer: \(\displaystyle \angle APB=\angle APC=60^\circ\), so \(\displaystyle PA\) bisects \(\displaystyle \angle BPC\).

  8. Exercise 8

    In Fig. 10.19\displaystyle 10.19, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠AEC=12\displaystyle \angle \mathrm{AEC}=\frac{1}{2} (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre). NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-4_Q8

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    Join \(\displaystyle AD\). NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q8 \[\angle DAB=\tfrac12\,\angle DOB \quad \text{(angle at centre = twice at circumference)} \] \[\angle ADC=\tfrac12\,\angle AOC \quad \text{(angle at centre = twice at circumference)} \] In \(\displaystyle \triangle ADE\), side \(\displaystyle DE\) is produced to \(\displaystyle C\), so \(\displaystyle \angle AEC\) is its exterior angle: \[\angle AEC=\angle EAD+\angle EDA \quad \text{(exterior angle of a triangle)} \] \[\angle AEC=\angle DAB+\angle ADC \] \[\Rightarrow \angle AEC=\tfrac12\,\angle DOB+\tfrac12\,\angle AOC =\tfrac12\left(\angle AOC+\angle DOB\right) \] Answer: \(\displaystyle \angle AEC=\tfrac12\left(\angle AOC+\angle DOB\right) \)

  9. Exercise 9

    If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle AP\) bisect \(\displaystyle \angle A\) and \(\displaystyle CQ\) bisect \(\displaystyle \angle C\), meeting the circle again at \(\displaystyle P, Q\). NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q9 \[\angle DAP=\angle PAB \ \Rightarrow\ \overset{\frown}{DP}=\overset{\frown}{PB} \quad \text{(equal angles subtend equal arcs)} \] \[\angle BCQ=\angle QCD \ \Rightarrow\ \overset{\frown}{BQ}=\overset{\frown}{QD} \quad \text{(equal angles subtend equal arcs)} \] \[\overset{\frown}{DP}+\overset{\frown}{PB}+\overset{\frown}{BQ}+\overset{\frown}{QD}=360^\circ \] \[2\overset{\frown}{PB}+2\overset{\frown}{BQ}=360^\circ \ \Rightarrow\ \overset{\frown}{PB}+\overset{\frown}{BQ}=180^\circ \] Answer: arc \(\displaystyle PBQ=180^\circ\), so \(\displaystyle PQ\) is a diameter.

  10. Exercise 10

    A circle has radius 2 cm\displaystyle \sqrt{2} \mathrm{~cm}. It is divided into two segments by a chord of length 2\displaystyle 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45\displaystyle 45°.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q10 \[OA=OB=\sqrt2\ \text{cm},\qquad AB=2\ \text{cm} \] \[OA^2+OB^2=2+2=4=AB^2 \] \[\Rightarrow \angle AOB=90^\circ \quad \text{(converse of Pythagoras theorem)} \] \[\angle APB=\tfrac12\,\angle AOB \quad \text{(angle at centre is twice angle at circumference)} \] \[\angle APB=45^\circ \]Answer: \(\displaystyle \angle APB = 45^\circ \).