Examples For Special Parallelograms

SOLVED EXAMPLES 1. In figure 7-34, the bisector of the angles of a parallelogram ABCD enclose quadrilateral PQRS. Prove that PQRS is a rectangle.

Sol: In parallelogram ABCD, ∠A + ∠B = 180o (Since Consecutive interior angles are supplementary)

1 Multiplying both sides of the above equation by , we get 2 1 1 (∠A + ∠B) = ×180o 2 2 or

1 1 ∠A + ∠B = 90o 2 2

or ∠PAB + ∠PBA = 90o In Triangle PAB, we have

… (1)

∠PAB + ∠PBA + ∠APB = 180o or 90o + ∠APB = 180o

[ Since ∠PAB + ∠PBA = 90o ]

or ∠APB = 180o − 90o = 90o or ∠P = 90o

[By using (1)]

… (2)

Again, in parallelogram ABCD ∠B + ∠C = 180o

(Consecutive interior angles are supplementary)

1 Multiplying both sides by , we get 2

1 1 (∠B + ∠C ) = ×180o 2 2 1 1 ∠B + ∠C ) = 90o 2 2 or ∠QBC + ∠QCB = 90o

… (3)

In Triangle QBC , ∠QBC + ∠QCB + ∠BQC = 180o (Angle sum property of triangle) or 90o + ∠BQC = 180o

[ Since ∠QBC + ∠QCB = 90o ]

or ∠BQC = 180o − 90o = 90o or ∠PQR = 90o

(Vertically opposite angles are equal)

or ∠Q = 90o

… (4)

Similarly, by using triangles formed on side CD and DA, we can show that

∠R = 90o

… (5)

and ∠S = 90o

… (6)

Thus in quadrilateral PQRS, we have

∠P = ∠Q = ∠R = ∠S = 90o

[By using (2), (4), (5) and (6)]

Hence PQRS is a rectangle 2. The diagonals of a rectangle ABCD meet at O. If ∠BOC = 40o . Find ∠OAD . Sol:

In figure 7-35, we have ∠DOA = ∠BOC

(Vertically opposite angles are equal)

or ∠DOA = 40o

… (1)

We know that the diagonals of a rectangle are equal and bisect each other Therefore AC=DB or

1 1 AC = DB 2 2

or OA=OD Now, in Triangle OAD

… (2)

OA=OD

[From (2)]

Therefore Triangle OAD is an isosceles triangle Thus, ∠OAD = ∠ODA

(in an isosceles triangle angles opposite equal sides are equal)

Now, in Triangle OAD ,

∠OAD + ∠ODA + ∠AOD = 180o

(Angle sum property of triangles)

or ∠OAD + ∠OAD + 40o = 180o

[ Since ∠OAD = ∠ODA and using (1)]

or 2∠OAD + 40o = 180o or 2∠OAD = 180o − 40o = 140o

140o or 2∠OAD = = 70o 2 3. If two parallel lines are intersected by a transversal, then prove that the bisectors of the interior angles form a rectangle. Sol:

Given: Two parallel lines AB and CD and a transversal EF intersecting them at G and H respectively. GM, HM, GL and HN are the bisectors of the two pairs of interior angles. To prove: GMHN is a rectangle Proof: We have ∠AGH = ∠DHG or

1 1 ∠AGH = ∠DHG 2 2

(Multiplying both sides by

1 ) 2

or ∠1 = ∠2 Thus lines GM and HN are intersected by a transversal GH at G and H respectively such that pair of alternate interior angles are equal. or ∠1 = ∠2

Therefore GM parallel to HN

Similarly, it can be proved that GN parallel to HM So GMHN is a parallelogram Since AB paralell to CD and EF is a transversal.

Therefore ∠BGH + ∠DHG = 180o (Consecutive supplementary)

interior

angles

1 Multiplying both sides by , we have 2 1 1 1 ∠BGH + ∠DHG = × 180o 2 2 2 or ∠3 + ∠2 = 90o

…. (1)

But ∠3 + ∠2 + ∠GNH = 180o

(Angle sum property of triangles)

or 90o + ∠GNH = 180o

using (1)

or ∠GNH = 180o − 90o = 90o Thus, in the parallelogram GMHN, we have ∠GNH = 90o Hence, GMHN is a rectangle (Since A parallelogram whose on angle is a right angle is a rectangle) 4. ABCD is a square, Find ∠DCA Sol: ABCD is a square.

Therefore AD = DC and ∠ADC = 90o Now, in Triangle ADC , we have

are

AD=DC

∠1 = ∠2

(Angles opposite equal sides of an isosceles triangle are equal and AD=DC)

But ∠1 + ∠2 + ∠ADC = 180o (Angle sum property of triangle) Since ∠1 = ∠2 and ∠ADC = 90o

Therefore ∠1 + ∠1 + 90o = 180o or 2∠1 + 90o = 180o or 2∠1 = 180o − 90o = 90o or ∠1 =

90o = 45o 2

or ∠2 = 45o

( Since ∠1 = ∠2 )

5. Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a quadrilateral is a parallelogram.

Sol: In figure 7-38, let ABCD be a quadrilateral in which P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively. We have to show that PQRS is a parallelogram. Join AC

In Triangle ABC , P and Q are the mid points of the sides AB and BC respectively

Therefore PQ parallel to AC and PQ =

1 AC 2

(mid point property)

… (1) In Triangle ACD, S and R are the mid points of the sides AD and CD respectively

Therefore SR parallel to AC and SR = … (2) From (1) and (2), we get

PQ parallel to SR and PQ=SR Therefore PQRS is a parallelogram

1 AC 2

(mid point property)

(If one pair of opposite sides of a quadrilateral are parallel and equal, then it is a parallelogram) 6. ABCD is a rectangle with ∠BAC = 30o . Determine ∠DBC . Sol: Suppose the diagonals AC and BD intersect at O. Then, in Triangle OAB each other)

(Since Diagonals of a rectangle are equal and bisect

and ∠OAB = ∠OBA (Angles opposite equal sides of an isosceles triangle) or ∠BAC = ∠DBA Thus ∠DBA = 30o

Now in ∠ABC = 90o or ∠DBA + ∠DBC = 90o or 30o + ∠DBC = 90o

( Since ∠BAC = 30o , given)

or ∠DBC = 90o − 30o = 60o 7. In figure 7-40, ABCD is a rhombus and P,Q, R and S are the mid points of AB, AC, CD and DA respectively, prove that PQRS is a rectangle. Sol: In Triangle ABC , P is the mid point of AB and Q is the mid point of BC.

Therefore PQ paralell to AC and PQ =

1 AC 2

(mid point theorem)

… (1)

In Triangle ACD , S is the mid point of AD and R is the mid point of CD.

Therefore SR parallel to AC and SR =

1 AC 2

(mid point theorem)

… (2)

From (1) and (2), we get

PQ parallel to SR and PQ=SR

Thus PQRS is a parallelogram (If one pair of opposite sides of quadrilateral is equal and parallel, quadrilateral is a parallelogram)

Similarly, by taking triangles ABD and Triangle BCD , we can show that

PS parallel to BD and QR parallel to BD Since ABCD is a rhombus, its diagonals AC and BD must be perpendicular. As PQ and SR are parallel to AC and PS and QR are parallel to BD, we get

∠P = ∠Q = ∠R = ∠S = 90o Therefore PQRS is a rectangle.

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