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SEBA Class 9 Mathematics Chapter 10 Circles
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Circles
Chapter – 10
| Exercise 10.1 |
1. Fill in the blanks:
(i) The centre of a circle lies in ___________ of the circle. (exterior/inte-rior).
Ans: The centre of a circle lies in interior of the circle.
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in ____________ of the circle. (exteror/interior).
Ans: A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
(iii) The longest chord of a circle is a _____________ of the circle.
Ans: The longest chord of a circle is a diameter of the circle.
(iv) An arcis a ____________ when its ends are the ends of a diameter.
Ans: An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and ___________of the circle.
Ans: Segment of a circle is the region between an arc and the chord of the circle.
(vi) A circle divides the plane, on which it lies, in ___________ parts.
Ans: A circle divides the plane, on which it lies, in three parts.
2. Write True or False: Give reasons for your answers.
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
Ans: True because all points on the circle equidistant from its centre.
(ii) A circle has only finite number of equal chords.
Ans: False because there are infinitely many points on the circle. So for each point on the circle, a point can be determined on the circle at a given distance from that point resulting into greatly many equal chords.
(iii) If a circle is divided into three equal arcs each is a major arc.
Ans: False because for each arc, the remaining arc will have greater length.
(iv) A chord of a circle, which is twice as long as its radius is a diameter of the circ.e
Ans: True because of definition of diameter.
(v) Sector is the region between the chord and its corresponding are.
Ans: False by virtue of its definition.
(vi) A circle is a plane figure.
Ans: True as it is a part of a plane.
| IMPORTANT POINTS |
1. Angle Subtended by a Chord at a Point
Take a line segments PQ and a point R not on the line con- taining PQ. Join PR and QR (see figure). Then∠PRQ is called the angle subtended by the line segment PQ at the point R.
In the figure, ∠POQ is the angle subtended by the chord PQ at the cen- tre O, ∠PRQ and ∠PSQ are respec-tively the angles subtended by PQ at points R and S on the major and minor arcs PQ respecively.
2. Relation between the Size of the Chord and the angle Subtended by it at the Centre:
Actively: Draw different chords of a circle and angles subtended by them at the centre. We see that the longer is the chord, the bigger will be the angle subtended by it at the centre.
Draw two or more chords of a circle and measure the angles subtended by them at the centre. We find that the angles subtended by them at the centre are equal.
3. Theorem
Equal chords of a circle subtended equal angles at the centre.
Given: AB and CD are the two equal chords of a circle with centre O.
To Prove: ∠AOB = ∠COD
Proof: In ∆AOB and ∆ COD
OA = OC | Radii of a circle
OB = OD | Radii of a circle
AB = CD | Given
∴ ∆AOB ≅ ∆COD | SSS Rule
∴ ∠AOB ≅ ∠ COD | c.p.c.t.
4. Converse of above Theorem
If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
Verification by activity
Take a tracing paper and trace a circle a circle on it. Cut it along the circle to get a disc. At its cen-tre O, draw an angle AOB where A, B are points on the circle. Make another angle POQ at the centre equal to ∠ AOB. Cut the disc along the ends of these angles (see figure). We will get two segments ACB and PRQ of the circle. If we put one on the other, we observe that they cover each other, i.e, they are congruent. So, AB = PQ.
Repeat this activity for other equal angles too. We observe that the chords turn out to be equal for any measure of equal angles.
Verification of exhaustion
Given: ∠ AOB and ∠ COD are two equal angles subtended by chords AB and CD of a circle at its centre O.
To Prove: AB = CD
Proof: In ∆AOB and ∆ COD
OA = OC | Radii of a circle
OB = OD | Radii of a circle
∠AOB = ∠COD | Given
∴ ∆AOB = ∆COD | SSS Rule
∴ AB = CD | c.p.c.t.
| Exercise 10.2 |
1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of a congruent circle subtend equal angles at their centres.
Ans: Given: AB and CD are two equal chords of congruent circles with centres O and O’ respectively.
To Prove: ∠AOB = ∠CO’D
Proof: In ∆AOB and ∆ O’CD,
OA = O’C | Radii of congruent circles
OB = O’D | Radii of congruent circles
AB = CD | Given
∴ ∆ OAB ≅ O’CD | SSS Rule
∴ ∠AOB ≅ ∠CO’D | c.p.c.t.
∴ ∠AOB = ∠CO’D
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Ans: ∠ AOB and ∠ CO’D are the two equal angles subtended by the chords AB and CD of two congruent circles with centres O and O’ respec-tively.
To Prove: AB = CD
Prove: In ΔΟΑΒ and ∆O’CD,
OA = O’C | Radii of congruent circles
OB = O’D | Radii of congruent circles
∠AOB = ∠CO’D | Given
∴ ΔΟΑΒ ≅ O’CD | SAS Rule
∴ AB = CD | c.p.c.t.
| IMPORTANTS |
1. Theorem
The perpendicular from the centre of a circle to a chord bisects the chord.
Verification by activity
Activity: Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Fold the paper along a line through O so that a portion of the chords falls on the other. Let the crease cut AB at the point M. Then,
∠OMA = ∠OMB = 90⁰ or OM is perpendicular to AB. Then point B coincide with A (see figure)
We observe that MA = MB
Verification by exhaustion
Given: A circle with centre O. AB is a chord of this circle. OM⊥AB
To Prove: MA = MB
Construction: Join OA and OB
Proof: In right triangles OMA and OMB,
OA = OB | Radii of a circle
OM = OM | Common
∴ ΔΟΜΑ ≅ ΔΟΜΒ | R.H.S. Rule
∴ MA = MB | c.p.c.t.
2. Converse of above Theorem
The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Given: A circle with centre O. AB is a chord of this circle whose midpoint is M.
To Prove: OM⊥AB
Construction: Join OA and OB
Proof: In ∆ΟΜΑ and ∆OMB,
MA = MB | Given
OM = OM | Common
OA = OB | Radii of a circle
∴ ∆OMA ≅ ∆ OMB | SSS Rule
∴ ∠AMO =∠BMO | c.p.c.t.
But ∠AMO + ∠BMO = 180⁰ | Linear Pair Axiom
∴ ∠AMO = ∠BMO = 90⁰
⇒ OM⊥AB
3. Theorem
There is one and only one circle passing through three given non-collinear points.
Activity: Take a point P. We see that there may be as many circles as we like passing through this point.
Now take two points P and Q. We again see that there may be an infinite number of circles passing through P and Q.
Now take three points A, B and C. Then, two cases arise:
Case I: Points A, B and C are collinear If the three points lie on a line, then the third point will lie inside or outside the circle passing through two points.
Case II: Points A, B and C Are not collinear Let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [see figure (i)]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively. Let these perpendicu-lar bisectors intersect at one point O. (Note that PQ and RS will intersect because they are not parallel) [see figure (ii)].
∵ O lies on the perpendicular bisector PQ of AB.
∴ OA = OB
∵ Every point on the perpendicular bisector of a line segment is equidistant from its end points
Similarly, ∵ O lies on the perpendicular bisector Rs of BC
∴ OB = OC
∵ Every point on the perpendicular bisector of a line segment is equidistant from its end points
So, OA = OB = OC
i.e, the points A, B and C are at equal distances from the point O.
So, if we draw a circle with centre O and radius OA it will also pass through B and C. This shows that there is a circle passing through the three points A, B and C. We know that two lines (perpendicular bisectors) can intersect at only one point, so we can draw only one circle with radius OA. In other words, there is a unique circle passing through A, B and C.
This proves the theorem.
Remark: If ABC is a triangle, then by the above theorem, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is called the circumcircle of the AABC. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.
| Exercise 10.3 |
1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Ans: Each pair has at the most two common points.
(i)
No point common
(ii)
One point common
(iii)
Two points common
The maximum number of common points is Two.
2. Suppose you are given a circle. Give a construction to find its centre.
Ans: Steps of Construction:
(i) Take any three points P, Q and R on the circle.
(ii) Join PQ and QR.
(iii) Draw the perpendicular bisectors PQ and QR. Let these intersect at O.
Then, O is the centre of the circle.
3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Ans: Given: Two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Prove: OP is the perpendicular bisector of AB.
Construction: Join OA, OB, PA and PB. Let OP intersect AB at M.
Proof: In ∆OAP and ∆OPB,
OA = OB | Radii of a circle
PA = PB | Radii of a circle
OP = OP | Common
∴ ∆ OAP ≅ ∆ OBP | SAS Rule
∴ ∠ AOP ≅ ∠BOP | c.p.c.
⇒ ∠AOM ≅ ∠BOM …(1)
In ∆AOM and ∆ BOM,
OA = OB | Radii of a circle
∠AOM =∠BOM | From (1)
OM = OM | Common
∴ ∆ AOM ≅ ∆BOM | SAS Rule
∴ AM = BM …(2) | c.p.c.t.
and ∠ AMO = ∠BMO …(3) | c.p.c.t.
But ∠AMO+∠BMO = 180⁰ | Linear Pair Axiom
∴ ∠AMO =∠BMO = 90⁰ …(4)
∴ OM, i.e., OP is the perpendicular bisector of AB. | From (2) and (4)
| Exercise 10.4 |
1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Ans: We know that if two circles in-tersect each other at two points, then line joining their centres is the per-ticular bisector of their common chord.
∴ Length of the common chord
= PQ = 2O’P
= 2 x 3cm = 6 cm
2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Ans: Given: A circle with centre O.
Its two equal chords AB and CD inter- sect at E.
To Prove: AE = DE and CE = BE
Construction: Draw ON⊥CD. Join OE
Proof: In ∆OME and ∆ONE,
OM = ON
| ∵ Equal chords of a circle are equidistant from the centre
OE = OE | Common
∴ ∆ OME ≅ ∆ONE | R.H.S.
∴ ME = NE | c.p.c.t.
3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Ans: Given: Two equal chords AB and CD of a circle with centre O inter-sect within the circle. Their point of in-tersection is E.
To Prove: ∠OEA = ∠OED
Construction: Join OA and OD.
Proof: In ∆OEA and ∆OED,
OE = OE | Common
OA = OD | Radii of a circle
AE = DE | Proved in Example 2 above
∴ ∆ OEA cong ∆ OED | SSS Rule
∴ ∠ OEA = ∠OED | c.p.c
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D. Prove that AB = CD [see figure]
Ans: Given: A line intersects two concentric circles (circles with the same centre) with cen-tre O at A, B, C and D.
To Prove: AB = CD
Construction: Draw OM⊥BC
Proof: ∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
∴ AM = DM …(1)
and BM = CM …(2)
Subtracting (2) from (1), we get
AM-BM-DM-CM
⇒ AB = CD
5. There girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma and Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Ans: In ∆NOR and ∆NOM,
ON = ON | Common
∠NOR = ΔΝΟΜ
| ∵ Equal chords of a circle subtend equal angles at the centre
OR = OM | Radii of a circle
∴ ∆NOR ≅ ∆NOM | SAS Rule
∴ ∠ONR = ∠ONM | c.p.c.t.
and NR = NM | c.p.c.t.
But ∠ONR+∠ONM = 180⁰ | Linear Pair Axiom
∴ ∠ONR = ∠ONM = 90⁰
∴ ON is the perpendicular bisector of RM
Draw bisector SN of ∠RSM to intersect the chord RM in N.
In ∆RSM and ∆MSN,
RS = MS (= 6 cm each)
SN=SN | Common
∠RSN = ∠MSN | By construction
∴ ∆RSN ≅ AMSN | SAS Rule
∴ ∠RNS = ∠MNS | c.p.c.t.
and RN = MN | c.p.c.t.
But ∠RNS +∠MNS = 180⁰ | Linear Pair Axiom
∴ ∠RNS = ∠MNS = 90⁰
∴ SN is the perpendicular bisector of RM and therefore passes through O when produced.
Let ON = xm
Then SN = (5 – x)m
In right triangle ONR,
x²+ RN² = 5² …(1) | By Pythagoras Theorem In right triangle SNR,
(5 – x)² + RN² = 6² …(2) | By Pythagoras Theorem
From (1),
RN² = 5²- x²
From (2),
RN² = 6²- (5 – x)²
Equating the two values of RN² ,we get
5² – x² = 6² – (5 – x)²
⇒ 25-x² = 36-(25-10x+x)² ⇒ 25 – x² = 36 – 25 + 10x – x²
⇒ 25 – x² = 11+10x-x² ⇒ 25-11 = 10x
⇒ 14 = 10x ⇒ 10 x = 14
Put x = 1.4 in (1), we get
(1.4)² + RN² = 5²
⇒ RN² = 5² – (1.4)²
⇒ RN² = 25 – 1.96
⇒ RN² = 23.04
⇒ RN = √(23.04)
⇒ RN = 4.8
∴ RM = 2RN = 2×4.8m = 9.6m
Hence the distance between Reshma and Mandip is 9.6 m.
6 A circular part of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Ans: Length of the string of each phone = Given equal distance.
7. The sum of the two opposite angles of a cyclic quadrilateral.
(a) 180⁰
(b) 110⁰
(c) 90⁰
(d) 360⁰
Ans: (a) 180⁰
| Exercise 10.5 |
1. In figure, A, B and C are three points on a circle with centre O such that ∠ BOC = 30⁰ and ∠ AOB = 60⁰ If D is a point on the circle other than the arc ABC, find ∠ ADC.
Ans:
| The angle subtended by an: the angle subtended by it at any point on the remaining part of the circle
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.
Ans: ∵ OA = OB = AB | Given
∴ ∆AOB is equilateral
∴ ∠ AOB = 60⁰
| The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
Now, ∵ ADBC is a cyclic quadrilateral.
∴ ∠ADB + ∠ACB = 180⁰
| The sum of either pair of opposite angles of a cyclic quadrilateral is 180⁰
⇒ ∠ADB + 30⁰ = 180⁰
⇒ ∠ADB = 180⁰- 30⁰
⇒ ∠ADB = 150⁰
3. In figure, ∠ PQR = 100⁰ where P, Q and R are points on a circle with centre O. Find ∠OPR.
Ans: Take a point S in the major arc. Join PS and RS.
∵ PQRS is a cyclic quadrilateral.
∵ ∠ PQR + ∠PSQ = 180⁰
| The sum of either pair of opposite angles of a cyclic quadrilateral is 180⁰
⇒ 100⁰ + ∠PSR = 180⁰
⇒ ∠ PSR = 180⁰ – 100⁰
⇒ ∠ PSR = 80⁰ …(1)
Now, ∠POR = 2∠PSR = 2 × 80⁰ = 160⁰ … (2) | Using (1)
| The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
In, DOPR
∵ OP = OR | Radii of a circle
∴ ∠OPR = ∠ORP …(3)
| Angles opposite to equal sides of a triangle are equal
In ∆OPR,
∠OPR + ∠ORP + ∠POR = 180⁰
Sum of all the angles of a triangle is 180⁰
⇒ ∠OPR + ∠OPR + 160 = 180⁰ | Using (2) and (1)
⇒ 2 ∠OPR + 160⁰ = 180⁰
⇒ 2 ∠OPR = 180⁰ – 160⁰ = 20⁰
⇒ ∠OPR = 10⁰
4. In figure,
∠ABC = 69⁰, ∠ACB = 31⁰ find ∠BDC
Ans: In ∆ ABC,
⇒ ∠ BAC + ∠ ABC + ∠ ACB = 180⁰ | Sum of all the angles of a triangle is 180⁰
⇒ ∠ BAC + 100⁰ = 180⁰
⇒ ∠ BAC = 180⁰ – 100⁰ = 80⁰
Now, ∠ BDC = ∠ BAC = 80⁰ | Angles in the same segment of a circle are equal
5. In figure, A, B, C and D are four points on a circle. AC and BD in-tersect at a point E such that ∠ BEC = 130⁰ and ∠ BCD = 20⁰ Find ∠ BAC.
Ans: ∠CED + ∠BEC = 180⁰ | Linear Pair Axiom
⇒ ∠CED + 130⁰ = 180⁰
⇒ ∠CED = 180⁰- 130⁰ = 50⁰ …(1)
∠ECD = 20⁰ …(2)
In ∆CED,
∠CED + ∠ECD + ∠CDE = 180⁰ | Sum of all the angles of a triangle is 180⁰
⇒ 50⁰ + 20⁰ + ∠CDE = 180⁰ | Using (1) and (2)
⇒ 70⁰+∠CDE = 180⁰
⇒ ∠CDE = 180⁰ – 130⁰ = 50⁰ …(1)
∠ECD = 20⁰ …(2)
In ∆CED,
∠CED + ∠ECD + ∠CDE = 180⁰ | Sum of all the angles of a triangle is 180⁰
⇒ 50⁰ + 20⁰ + ∠CDE = 180⁰ | Using (1) and (2)
⇒ 70⁰ + ∠CDE = 180⁰
⇒ ∠CDE = 180⁰ – 70⁰
⇒ ∠CDE = 110⁰
Now, ∠BAC = ∠CDE | Angles in the same segment of a circle are equal
= 110⁰ | Using (3)
6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If angle DBC = 70⁰, ∠BAC is 30⁰, find ∠BCD. Further, if AB = BC, find ∠ECD.
Ans: ∠CDB = ∠BAC | Angles in the same segment of a circle are equal
= 30° …(1)
∠DBC = 70⁰ …(2)
In ∆BCD,
∠BCD + ∠BDC + ∠CDB = 180⁰ | Sum of the angles of triangle is 180⁰
⇒ ∠BCD + 70⁰ + 30⁰ = 180⁰ | Using (1) and (2)
⇒ ∠BCD + 100⁰ = 180⁰
⇒ ∠BCD = 180⁰ – 100⁰
⇒ ∠BCD = 80⁰ …(3)
In ∆ABC, AB = BC
∴ ∠BCA = ∠BAC | Angles opposite to equal sides of A a triangle are equal
= 30⁰ …(4) | ∵ ∠BAC=30⁰ (given)
Now, ∠BCD = 80⁰ | From (3)
⇒ ∠BCA + ∠ECD = 80⁰
⇒ 30⁰ + ∠ECD = 80⁰
⇒ ∠ECD = 80⁰ – 30⁰
⇒ ∠ECD = 50⁰
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Ans: In and ∆OCD,
OA = OC | Radii of a circle
OB = OD | Radii of a circle
∠AOB = ∠COD | Vertically Opposite Angles
∴ ΔΑΟΒ ≅ ∆OCD | SAS Rule
∴ AB = CD | c.p.c.t.
⇒ Arc AB = Arc CD …(1)
Similarly, we can show that
Arc AD = Arc CB …(2)
Adding (1) and (2), we get
Arc AB + Arc AD + Arc CB
⇒ Arc BAD = Arc BCD
⇒ BD divides the circle into two equal parts (each semicircle)
∴ ∠A = 90⁰ ∠C = 90⁰ | Angle of a semicircle is 90⁰
Similarly, we can show that
∠B = 90⁰, ∠D=90⁰
∴ ∠A = ∠B = ∠C = ∠D = 90⁰
∴ ABCD is a rectangle.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Ans: Given: ABCD is cyclic.
Construction: Draw BE || AD
Proof: ∵ AB || DE | Given
AB || BE | By construction
∴ Quadrilateral ABCD is a parallelogram.
∴ ∠BAD = ∠BED …(1) | Opp. ∠s of all || gm
and AD = BE …(2) | Opp. sides of a || gm
But AD = BC …(3) | Given
From (2) and (3), BE = BC
∴ ∠BEC = ∠BCE …(4) | Angles opposite to equal sides
∠BEC + ∠BED = 180⁰ | Linear Pair Axiom
⇒ ∠BCE + ∠BAD = 180⁰ | From (4) and (1)
⇒ Trapezium ABCD is cyclic.
∵ If a pair of opposite angles of a quadrilateral is 180⁰, then the quadrilateral is cyclic.
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 figure). Prove that ∠ACP = ∠QCD.
Ans: Given: Two circles in-tersect at two points B and C.
Through B. two line segments ABD and PBQ are drawn to in- tersect the circles at A, D and P,Q respectively.
To Prove: ∠ACP = ∠QCD
Proof: ∠ACP = ∠ABP…(1) | Angles in the same segment of a circle are equal
∠QCD = ∠QBD…(2) | Angles in the same seg- ment of a circle are equal
∠ABD = ∠QBD…(3) | Vertically Opposite Angles
From (1), (2) and (3), ∠ACP = ∠QCD
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.
Ans: Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect in a point D.
To Prove: D lies on the third side BC of ∆АВС.
Construction: Join AD.
Proof: ∵ Circle described on AB as diameter intersects BC in D.
∴ ∠ADB = 90⁰ | Angle in a semi-circle
But ∠ADB + ∠ADC = 180⁰ | Linear Pair Axiom
∴ ∠ADC = 90⁰
Hence, the circle described on AC as diameter must pass through D.
Thus, the two circles intersect in D.
Now, ∠ADC = 180⁰
∴ Points B, D, C are collinear.
∴ D lies on BC.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD
Ans: Given: ABC and ADC are two right triangles with common hypotenuse AC.
To Prove: ∠CAD = ∠CBD
Proof: ∵ AC is the common hypotenuse and ABC and ADC are two right triangles.
∴ ∠ABC = 90⁰ = ∠ADC
⇒ Both the triangles are in the same semi-circle.
∴ Points A, B, D and C are concyclic.
DC is a chord
∴ ∠CAD = ∠CBD | ∵ Angles in the same segment are equal
12. Prove that a cyclic parallelogram is a rectangle.
Ans: Given: ABCD is a cyclic parallelogram.
To Prove: ABCD is a rectangle.
Proof: ∵ ABCD is a cyclic quadrilateral
∴ ∠1 + ∠2 = 180⁰ …(1) | ∵ Opposite angles of a cyclic quadrilateral are supplementary
∴ ABCD is a parallelogram
∴ ∠1 + ∠2 = …(2) | Opp. angles of a || gm
From (1) and (2), ∠1 = ∠2 = 90⁰
∴ || gm ABCD is a rectangle.
