Rectilinear Figures — Exercise 12.1

If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.

Answer

Sum of angles of quadrilateral = 360°.

Let other two angles be 3x and 4x.

∴ 3x + 4x + 40° + 110° = 360°

⇒ 7x + 150° = 360°

⇒ 7x = 360° - 150°

⇒ 7x = 210°

⇒ x = $\dfrac{210°}{7}$

⇒ x = 30°

⇒ 3x = 3(30°) = 90° and 4x = 4(30°) = 120°.

Hence, other two angles are 90° and 120°.

If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.

Answer

Let a trapezium be ABCD.

Let angles be x, 2x, 3x and 4x taken in order.

Sum of angles of quadrilateral = 360°.

∴ x + 2x + 3x + 4x = 360°

⇒ 10x = 360°

⇒ x = $\dfrac{360°}{10}$

⇒ x = 36°.

⇒ ∠A = x = 36°,

⇒ ∠B = 2x = 2(36°) = 72°

⇒ ∠C = 3x = 3(36°) = 108°

⇒ ∠D = 2x = 4(36°) = 144°.

From figure,

∠A and ∠D are co-interior angles and their sum = 180° (36° + 144°).

Sum of co-interior angles = 180°, which means AB and CD are parallel.

Also,

Opposite angles sum i.e.

∠A + ∠C = 36° + 108° = 144°,

∠B + ∠D = 72° + 144° = 216°.

Since, the sum of opposite angles ≠ 180° (Property of trapezium).

Hence, proved that ABCD is a trapezium.

If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.

Answer

Let a parallelogram be ABCD.

Let ∠A = x and so adjacent angle (∠B) = $\dfrac{2}{3}x$

As AD || BC, sum of co-int ∠s = 180°,

$\therefore x + \dfrac{2}{3}x = 180° \\[1em] \Rightarrow \dfrac{3x + 2x}{3} = 180° \\[1em] \Rightarrow \dfrac{5x}{3} = 180° \\[1em] \Rightarrow x = \dfrac{180° \times 3}{5} \\[1em] \Rightarrow x = 108° \\[1em] \Rightarrow \dfrac{2x}{3} = \dfrac{2 \times 108°}{3} = 72°.$

Opposite angles of a parallelogram are equal.

∴ ∠C = ∠A = 108° and ∠D = ∠B = 72°

Hence, angles of parallelogram are 108°, 72°, 108° and 72°.

In the figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.

Answer

As AD || BC, sum of co-int ∠s = 180°.

∠A + ∠B = 180°

∠B = 180° - ∠A = 180° - 70° = 110°.

From figure,

⇒ ∠B = ∠DBA + ∠DBC

⇒ 110° = ∠DBA + 80°

⇒ ∠DBA = 110° - 80° = 30°.

Sum of angles of a triangle = 180°

In △DAB,

⇒ ∠DAB + ∠DBA + ∠ADB = 180°

⇒ 70° + 30° + ∠ADB = 180°

⇒ ∠ADB = 180° - 100° = 80°.

Since, opposite angles of a parallelogram are equal,

∴ ∠C = ∠A = 70°.

In △DBC,

⇒ ∠DBC + ∠BCD + ∠CDB = 180°

⇒ 80° + 70° + ∠CDB = 180°

⇒ ∠CDB = 180° - 150° = 30°.

Hence, ∠ADB = 80° and ∠CDB = 30°.

In figure (2) given below, ABCD is a parallelogram. Find the angles of △AOD.

Answer

Sum of angles of a triangle = 180°

In △BOC,

⇒ ∠BOC + ∠OCB + ∠CBO = 180°

⇒ ∠BOC + 35° + 77° = 180°

⇒ ∠BOC = 180° - 35° - 77° = 68°.

From figure,

∠AOD = ∠BOC = 68° (Vertically opposite angles are equal)

∠OAD = ∠OCB = 35° (Alternate angles are equal)

∠ADO = ∠CBO = 77° (Alternate angles are equal)

Hence, ∠AOD = 68°, ∠OAD = 35°, ∠ADO = 77°.

In figure (3) given below, ABCD is a rhombus. Find the value of x.

Answer

As AD || BC, sum of co-int ∠s = 180°.

∠A + ∠B = 180°

∠B = 180° - ∠A = 180° - 72° = 108°.

The diagonals of rhombus bisect the angles.

$\therefore x° = \dfrac{1}{2} \times 108° = 54°.$

⇒ x = 54.

Hence, x = 54.

In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.

Answer

In parallelogram opposite sides are equal.

∴ BC = AD = 2x and

AB = CD

⇒ 3x = 2y + 2

⇒ 2y = 3x - 2

⇒ y = $\dfrac{3x - 2}{2}$ ........(i)

Given, perimeter = 40.

⇒ AB + BC + CD + AD = 40

⇒ 3x + 2x + 2y + 2 + 2x = 40

⇒ 3x + 2x + $2\Big(\dfrac{3x - 2}{2}\Big)$ + 2 + 2x = 40

⇒ 3x + 2x + 3x - 2 + 2 + 2x = 40

⇒ 10x = 40

⇒ x = 4.

⇒ y = $\dfrac{3x - 2}{2} = \dfrac{3 \times 4 - 2}{2} = \dfrac{12 - 2}{2} = 5$.

Hence, x = 4 and y = 5.

In figure (2) given below, ABCD is a parallelogram. Find the values of x and y.

Answer

In parallelogram,

Opposite angles are equal

∴ ∠A = ∠C

⇒ 3x - 20° = x + 40°

⇒ 3x - x = 40° + 20°

⇒ 2x = 60°

⇒ x = 30°.

As AD || BC, sum of co-int ∠s = 180°.

⇒ ∠A + ∠B = 180°

⇒ 3x - 20° + y + 15° = 180°

⇒ 3(30°) - 20° + y + 15° = 180°

⇒ 90° - 20° + y + 15° = 180°

⇒ 85° + y = 180°

⇒ y = 95°.

Hence, x = 30° and y = 95°.

In figure (3) given below, ABCD is a rhombus. Find x and y.

Answer

Sides of rhombus are equal.

∴ AB = AD

⇒ 4x - 4 = 3x + 2

⇒ 4x - 3x = 2 + 4

⇒ x = 6.

AB = 3x + 2 = 3(6) + 2 = 20.

In △ABD,

AB = AD

so, ∠D = ∠B (Angles opposite to equal sides are equal in isosceles triangle)

Let angle be ∠D = ∠B = a

So,

⇒ ∠A + ∠D + ∠B = 180°

⇒ 60° + a + a = 180°

⇒ 2a = 180° - 60°

⇒ 2a = 120°

⇒ a = 60°.

Hence, △ABD is an equilateral triangle, so all sides are equal.

⇒ BD = AB

⇒ y - 1 = 20

⇒ y = 21.

Hence x = 6 and y = 21.

The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.

Answer

The diagonals of a rectangle are equal and bisect each other.

∴ AP = BP

From figure,

In △ABP,

AP = BP

⇒ ∠PAB = ∠ABP = 50° (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠PAB + ∠ABP + ∠APB = 180°

⇒ 50° + 50° + ∠APB = 180°

⇒ ∠APB = 180° - 100° = 80°.

⇒ ∠DPC = ∠APB = 80° (Vertically opposite angles are equal).

Hence, ∠DPC = 80°.

In figure (1) given below, equilateral triangle EBC surmounts square square ABCD. Find angle BED represented by x.

Answer

△EBC is an equilateral triangle, so all sides are equal.

EB = BC = EC ........(i)

In square all sides are equal

AD = CD = BC = AB ........(ii)

From (i) and (ii) we get,

BC = EC = CD

⇒ EC = CD.

In △ECD,

EC = CD

⇒ ∠DEC = ∠CDE = a (let) (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠C = ∠ECB + ∠BCD

∠ECB = 60° (As each angle of a equilateral triangle = 60°)

∠BCD = 90° (As each angle of a square = 90°)

⇒ ∠C = 60° + 90° = 150°.

⇒ ∠DEC + ∠CDE + ∠C = 180

⇒ a + a + 150° = 180°

⇒ 2a = 180° - 150°

⇒ 2a = 30°

⇒ a = 15°.

From figure,

x° = ∠BEC - ∠DEC = 60° - 15° = 45°.

Hence, x = 45.

In Figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of △AOB.

Answer

From figure,

∠OCD = 180° - 146° = 34° (As AE is a straight line).

The diagonals of a rectangle are equal and bisect each other.

∴ OC = OD

From figure,

In △OCD,

OC = OD

⇒ ∠ODC = ∠OCD = 34° (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠ODC + ∠OCD + ∠DOC = 180°

⇒ 34° + 34° + ∠DOC = 180°

⇒ ∠DOC = 180° - 68° = 112°.

In △AOB,

⇒ ∠AOB = ∠DOC = 112° (Vertically opposite angles are equal).

⇒ ∠OAB = ∠OCD = 34°

⇒ ∠OBA = ∠ODC = 34°

Hence, ∠AOB = 112°, ∠OAB = 34° and ∠OBA = 34°.

In figure (3) given below, ABCD is a rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3 : 2, find the angles of the △AOD.

Answer

Let ∠OAB = 3x and ∠OBA = 2x.

The diagonals of rhombus are perpendicular to each other.

∴ ∠AOB = 90°

In △AOB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ 90° + 3x + 2x = 180°

⇒ 5x = 90°

⇒ x = $\dfrac{90°}{5}$

⇒ x = 18°.

∠OAB = 3x = 3(18°) = 54°.

∠OBA = 2x = 2(18°) = 36°.

Since, diagonals of rhombus bisect vertex angles.

∴ ∠OAD = ∠OAB = 54°,

∠AOD = 90° (The diagonals of rhombus are perpendicular to each other.)

In △AOD,

⇒ ∠AOD + ∠OAD + ∠ODA = 180°

⇒ 90° + 54° + ∠ODA = 180°

⇒ ∠ODA + 144° = 180°

⇒ ∠ODA = 180° - 144°

⇒ ∠ODA = 36°.

Hence, ∠ODA = 36°, ∠OAD = 54°, ∠AOD = 90°.

In figure (1) given below, ABCD is a trapezium. Find the values of x and y.

Answer

Sum of adjacent co-interior angles of a trapezium = 180° (As AB || DC)

∴ ∠A + ∠D = 180°

⇒ x + 20° + 2x + 10° = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 150°

⇒ x = $\dfrac{150°}{3}$

⇒ x = 50°.

∴ ∠B + ∠C = 180°

⇒ 92° + y = 180°

⇒ y = 180° - 92° = 88°.

Hence, x = 50° and y = 88°.

In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and y.

Answer

Sum of adjacent co-interior angles of a trapezium = 180° (As AB || DC)

∴ ∠A + ∠D = 180°

⇒ 2x + 3x = 180°

⇒ 5x = 180°

⇒ x = $\dfrac{180°}{5}$

⇒ x = 36°.

Opposite angles sum of a isosceles trapezium = 180°

∴ ∠A + ∠C = 180°

⇒ 2x + y = 180°

⇒ 2(36°) + y = 180°

⇒ 72° + y = 180°

⇒ y = 108°.

Hence, x = 36° and y = 108°.

In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.

Answer

Diagonals of a kite bisect the vertex angles.

∴ ∠OCD = $\dfrac{∠C}{2} = \dfrac{64}{2}$ = 32° and ∠OAB = $\dfrac{∠A}{2} = \dfrac{112}{2}$ = 56°

Diagonals of a kite are perpendicular to each other.

∴ ∠DOC = 90°

⇒ ∠OCD + ∠DOC + ∠ODC = 180°

⇒ 32° + 90° + ∠ODC = 180°

⇒ ∠ODC = 58°.

Diagonals of a kite are perpendicular to each other.

∴ ∠AOB = 90°

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ 90° + 56° + ∠OBA = 180°

⇒ ∠OBA = 180° - 146° = 34°.

Hence, ∠OBA = 34° and ∠ODC = 58°.

Prove that each angle of a rectangle is 90°.

Answer

We know that opposite sides of a rectangle are equal.

∴ AD = BC and AB = CD

Also, the diagonals of rectangle are equal.

∴ AC = BD

Now, consider ΔADC and ΔBCD

⇒ AD = BC

⇒ AC = BD

⇒ DC = DC [Common]

∴ ΔADC ≅ ΔBCD by SSS congruency rule.

∴ ∠ ADC = ∠BCD = x (let) [By C.P.C.T.]

But, adjacent sides of a parallelogram are supplementary. [∵ rectangle is a parallelogram].

∴ ∠ADC + ∠BCD = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = 90°

⇒ ∠ADC = ∠BCD = 90°.

Since, opposite angles of a parallelogram are also equal.

∴ ∠DAB = ∠BCD = 90° and ∠ABC = ∠ADC = 90°.

Hence, proved that each angle of a rectangle is 90°.

If the angle of a quadrilateral are equal, prove that it is a rectangle.

Answer

Suppose there is a quadrilateral ABCD.

Let ∠A = ∠B = ∠C = ∠D = x

So, ∠A = ∠C and ∠B = ∠D (Opposite angles are equal)

∴ ABCD is a parallelogram.

Since, sum of angles in a quadrilateral = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ x + x + x + x = 360°

⇒ 4x = 360°

x = $\dfrac{360°}{4}$

⇒ x = 90°.

∴ ∠A = ∠B = ∠C = ∠D = 90°.

Since, each angle = 90°,

Hence, proved that ABCD is a rectangle.

If the diagonals of a rhombus are equal, prove that it is a square.

Answer

From figure,

In △ABC and △BCD,

AB = DC (All sides of rhombus are equal)

BC = BC (Common sides)

AC = BD (Given diagonals are equal)

∴ △ABC ≅ △BCD (By SSS rule of congruency)

∠ABC = ∠BCD = x (let) (By C.P.C.T.)

∠ABC + ∠DCB = 180° [∵ AB || DC, sum of co-int ∠s = 180°]

⇒ x + x = 180°

⇒ 2x = 180°

x = $\dfrac{180°}{2}$

⇒ x = 90°.

∴ ∠ABC = ∠DCB = 90°.

Hence, proved that ABCD is a square.

Prove that every diagonal of a rhombus bisects the angles at the vertices.

Answer

From figure,

In △AOD and △COD,

AD = CD (All sides of rhombus are equal)

DO = OD (Common sides)

AO = OC (Diagonals of rhombus bisect each other)

∴ △AOD ≅ △COD (By SSS rule of congruency)

∴ ∠ADO = ∠CDO (By C.P.C.T.)

∴ BD bisects ∠D.

In △AOB and △COB,

AB = BC (All sides of rhombus are equal)

BO = OB (Common sides)

AO = OC (Diagonals of rhombus bisect each other)

∴ △AOB ≅ △COB (By SSS rule of congruency)

∴ ∠ABO = ∠CBO (By C.P.C.T.)

∴ BD bisects ∠B.

In △AOB and △AOD,

AB = AD (All sides of rhombus are equal)

AO = OA (Common sides)

OD = OB (Diagonals of rhombus bisect each other)

∴ △AOB ≅ △AOD (By SSS rule of congruency)

∴ ∠OAD = ∠OAB (By C.P.C.T.)

∴ AC bisects ∠A.

In △BOC and △DOC,

BC = CD (All sides of rhombus are equal)

OC = CO (Common sides)

OD = OB (Diagonals of rhombus bisect each other)

∴ △BOC ≅ △DOC (By SSS rule of congruency)

∴ ∠OCD = ∠OCB (By C.P.C.T.)

∴ AC bisects ∠C.

Hence, proved that every diagonal of a rhombus bisects the angles at the vertices.

ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that :

(i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC ⊥ BD

Answer

Parallelogram ABCD is shown in the figure below:

(i) Given,

AC bisects ∠A,

∴ ∠CAB = ∠CAD = x (let) .......(i)

AB || CD (As opposite sides of parallelogram are parallel)

⇒ ∠DCA = ∠CAB (Alternate angles are equal)

∴ ∠DCA = x .......(ii)

and,

⇒ ∠BCA = ∠CAD (Alternate angles are equal)

∴ ∠BCA = x .......(iii)

From (i), (ii) and (iii) we get,

∠CAB = ∠CAD = ∠DCA = ∠BCA .......(iv)

Thus, ∠DCA = ∠BCA

Hence, proved that AC bisects ∠C.

(ii) From equation (iv) we get,

∠CAD = ∠DCA

∴ In △ADC,

DA = DC (Sides opposite to equal angles are equal)

However, DA = BC and AB = CD (opposite sides of a parallelogram are equal)

Thus, AB = BC = CD = DA

As ABCD is a parallelogram in which all sides are equal, therefore ABCD is a rhombus.

Hence, proved that ABCD is a rhombus.

(iii) In △OAB and △OCB,

OA = OC (Diagonals of a parallelogram bisect each other)

OB = OB (Common Side)

AB = BC (Sides of Rhombus)

∴ △OAB ≅ △OCB (SSS rule of congruency)

∴ ∠AOB = ∠BOC (c.p.c.t)

But ∠AOB + ∠BOC = 180°

⇒ ∠AOB + ∠AOB = 180°

⇒ ∠AOB = $\dfrac{180°}{2}$

⇒ ∠AOB = 90°

∴ AC ⊥ BD

Hence, proved that AC ⊥ BD.

Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.

Answer

Let AC be bisector of ∠A and BD be bisector of ∠B and they meet at point M.

From figure,

⇒ ∠A + ∠B = 180° (As AD || BC, sum of co-int ∠s = 180°)

⇒ $\dfrac{∠A + ∠B}{2} = \dfrac{180°}{2} = 90.$

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2} = 90°$

∴ ∠MAB + ∠MBA = 90° .......(i)

In △MAB,

⇒ ∠MAB + ∠MBA + ∠AMB = 180° (Sum of angles of triangle = 180°)

⇒ 90° + ∠AMB = 180° (from i)

⇒ ∠AMB = 180° - 90°

⇒ ∠AMB = 90°.

Hence, proved that bisectors of any two adjacent angles of a parallelogram are at right angles.

Prove that bisectors of any two opposite angles of a parallelogram are parallel.

Answer

Let the parallelogram be ABCD as shown in the figure below:

In parallelogram ABCD we have,

∠A = ∠C (Opposite angles are equal)

so,

$\dfrac{∠A}{2} = \dfrac{∠C}{2}$

∠DAR = ∠QCB (As AR bisects ∠A and QC bisects ∠C and ∠A = ∠C)

In △ADR and △CBQ,

⇒ ∠DAR = ∠QCB (Proved above)

⇒ AD = BC (Opposite sides of a || gm)

⇒ ∠D = ∠B (Opposite angles of a || gm)

Hence, △ADR ≅ △CBQ by ASA axiom.

∴ ∠DRA = ∠BQC (By C.P.C.T.) .......(i)

Also,

∠RAQ = ∠DRA (Alternate angles are equal) .........(ii)

From (i) and (ii) we get,

∠RAQ = ∠BQC (These are also corresponding angles)

Since, corresponding angles are equal, we can say that

AR || QC.

Hence, proved that bisectors of any two opposite angles of a parallelogram are parallel.

If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.

Answer

Since, diagonals bisect each other at 90°

∴ ∠AOB = ∠COD = ∠BOC = AOD = 90°.

From figure,

Considering △OAB and △ODC we have,

⇒ OA = OC (As diagonals bisect each other)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOB = ∠COD (Vertically opposite angles are equal)

Hence, △OAB ≅ △ODC by SAS axiom.

AB = CD (By C.P.C.T.) .........(i)

∴ ∠OAB = ∠OCD (By C.P.C.T.)

The above angles are alternate angles.

Hence, we can say that AB || CD.

In △AOB,

OA = OB (As both diagonals are equal and bisect each other)

so,

⇒ ∠OBA = ∠OAB = x (let) (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

Considering △OAD and △OBC we have,

⇒ OA = OC (As diagonals bisect each other)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOD = ∠COB (Vertically opposite angles are equal)

Hence, △OAD ≅ △OBC by SAS axiom.

AD = BC (By C.P.C.T.) .........(ii)

∠OAD = ∠OCB (By C.P.C.T.)

Hence, we can say that AD || BC.

Considering △AOB and △AOD we have,

⇒ AO = AO (Common sides)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOD = ∠AOB (Both equal to 90°)

Hence, △AOB ≅ △AOD by SAS axiom.

AB = AD (By C.P.C.T.) .........(iii)

From (i), (ii) and (iii) we get,

AB = BC = CD = AD.

In △AOD,

OA = OD (As both diagonals are equal and bisect each other)

so,

⇒ ∠OAD = ∠ODA = a (let) (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠OAD + ∠ODA + ∠AOD = 180°

⇒ a + a + 90° = 180°

⇒ 2a = 180° - 90°

⇒ 2a = 90°

⇒ a = $\dfrac{90°}{2}$

⇒ a = 45°.

∠A = ∠OAB + OAD = 45° + 45° = 90°.

Thus, AB ⊥ AD.

Since, AD || BC so, AB ⊥ BC.

Since, AB || CD and AB ⊥ AD,

∴ CD ⊥ AD.

Since, alternate sides are perpendicular and all sides are equal.

Hence, proved that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square

If ABCD is a parallelogram in which diagonal AC bisects ∠A, then prove that ABCD is a rhombus.

Answer

Parallelogram ABCD is shown in the figure below:

Given,

AC bisects ∠A,

∴ ∠CAB = ∠CAD = x (let) ....................(1)

AB || CD (As opposite sides of parallelogram are parallel)

⇒ ∠DCA = ∠CAB (Alternate angles are equal)

∴ ∠DCA = x ....................(2)

and,

⇒ ∠BCA = ∠CAD (Alternate angles are equal)

∴ ∠BCA = x ....................(3)

From (1), (2) and (3) we get,

⇒ ∠CAB = ∠CAD = ∠DCA = ∠BCA ....................(4)

⇒ ∠CAD = ∠DCA

∴ In △ADC,

⇒ ∠CAD = ∠DCA

⇒ DA = DC (Sides opposite to equal angles in a triangle are equal) .........(5)

We know that,

Opposite sides of a parallelogram are equal.

AB = CD and BC = DA ...........(6)

From equation (5) and (6),

AB = BC = CD = DA

As ABCD is a parallelogram in which all sides are equal, therefore ABCD is a rhombus.

Hence, proved that ABCD is a rhombus.

If ABCD is a rectangle in which the diagonal BD bisects ∠B, then show that ABCD is a square.

Answer

Since, BD bisects ∠B.

∴ ∠1 = ∠2

Since, ∠B = 90°.

∠1 = ∠2 = 45°

∠4 = ∠1 = 45° (Alternate angles are equal)

∠3 = ∠2 = 45° (Alternate angles are equal)

In △ABD,

AB = AD (As sides opposite to equal angles are equal) .........(i)

In △CBD,

BC = CD (As sides opposite to equal angles are equal) ..........(ii)

and AD = BC, AB = CD (As opposite sides of rectangle are equal) .......(iii)

From (i), (ii) and (iii) we get,

AB = BC = CD = AD.

Since, all sides are equal and alternate sides are perpendicular to each other.

Hence, proved that ABCD is a square.

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Answer

Since, diagonals bisect each other at 90°

∴ ∠AOB = ∠COD = BOC = AOD = 90°.

From figure,

Considering △OAB and △ODC we have,

⇒ OA = OC (As diagonals bisect each other)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOB = ∠COD (Vertically opposite angles are equal)

Hence, △OAB ≅ △ODC by SAS axiom.

AB = CD (By C.P.C.T.) .........(i)

∴ ∠OAB = ∠OCD (By C.P.C.T.)

The above angles are alternate angles.

Hence, we can say that AB || CD.

In △AOB,

OA = OB (As both diagonals are equal and bisect each other)

so,

⇒ ∠OBA = ∠OAB = x (let) (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°.

Considering △OAD and △OBC we have,

⇒ OA = OC (As diagonals bisect each other)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOD = ∠COB (Vertically opposite angles are equal)

Hence, △OAD ≅ △OBC by SAS axiom.

AD = BC (By C.P.C.T.) .........(ii)

∠OAD = ∠OCB (By C.P.C.T.)

Hence, we can say that AD || BC.

Considering △AOB and △AOD we have,

⇒ AO = AO (Common side)

⇒ OB = OD (As diagonals bisect each other)

⇒ ∠AOD = ∠AOB (Both equal to 90°)

Hence, △AOB ≅ △AOD by SAS axiom.

AB = AD (By C.P.C.T.) .........(iii)

From (i), (ii) and (iii) we get,

AB = BC = CD = AD.

In △AOD,

OA = OD (As both diagonals are equal and bisect each other)

so,

⇒ ∠OAD = ∠ODA = a (let) (Angles opposite to equal side are equal in isosceles triangle)

⇒ ∠OAD + ∠ODA + ∠AOD = 180°

⇒ a + a + 90° = 180°

⇒ 2a = 180° - 90°

⇒ 2a = 90°

⇒ a = $\dfrac{90°}{2}$

⇒ a = 45°.

∠A = ∠OAB + ∠OAD = 45° + 45° = 90°.

Thus, AB ⊥ AD.

Since, AD || BC so, AB ⊥ BC.

Since, AB || CD and AB ⊥ AD,

∴ CD ⊥ AD.

Since, alternate sides are perpendicular and all sides are equal.

Hence, proved that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square

P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Answer

Parallelogram ABCD is shown in the figure below:

Considering △OAP and △OCQ we have,

⇒ ∠OAP = ∠OCQ (Alternate angles are equal)

⇒ OA = OC (As diagonals bisect each other)

⇒ ∠AOP = ∠COQ (Vertically opposite angles)

Hence, △OAP ≅ △OCQ by ASA axiom.

OP = OQ (By C.P.C.T.)

Hence, proved that PQ is bisected at O.

In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that

(i) the triangles ABX and QCX are congruent.

(ii) DC = CQ = QP

Answer

(i) Considering △ABX and △QCX we have,

⇒ ∠XAB = ∠XQC (Alternate angles are equal)

⇒ XB = XC (As X is mid-point of BC)

⇒ ∠AXB = ∠CXQ (Vertically opposite angles are equal)

Hence, △ABX ≅ △QCX by ASA axiom.

(ii) Since, △ABX ≅ △QCX

∴ AB = CQ (By C.P.C.T.) ..........(i)

AB = CD and AB = QP (Opposite sides of parallelogram are equal) .........(ii)

From (i) and (ii) we get,

⇒ AB = DC = CQ = QP

⇒ DC = CQ = QP

Hence, proved that DC = CQ = QP.

In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

Answer

Considering △AOP and △COQ we have,

⇒ ∠OAP = ∠OCQ (Alternate angles are equal)

⇒ AP = QC (Given)

⇒ ∠AOP = ∠COQ (Vertically opposite angles are equal)

Hence, △AOP ≅ △COQ by ASA axiom.

∴ AO = OC and OP = OQ (By C.P.C.T.)

Hence, proved that AC and PQ bisect each other at point O.

ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.

Answer

Square ABCD is shown in the figure below:

Considering △ABP and △ADQ we have,

⇒ ∠ABP = ∠DAQ = 90°

⇒ AP = DQ (Given)

⇒ AB = AD (Sides of square are equal)

Hence, △ABP ≅ △ADQ by RHS axiom.

⇒ ∠BAP = ∠ADQ (By C.P.C.T.) ........(i)

⇒ ∠BAD = 90° (Each angle of square = 90°)

⇒ ∠BAP + ∠PAD = 90°

Substituting value of ∠ADQ from (i) we get,

⇒ ∠ADQ + ∠PAD = 90° .........(ii)

From figure,

∠ADQ = ∠ADM

∠PAD = ∠MAD

Substituting above values in (ii) we get,

⇒ ∠ADM + ∠MAD = 90° .......(iii)

In △AMD,

⇒ ∠ADM + ∠MAD + ∠AMD = 180°

⇒ 90° + ∠AMD = 180° (From iii)

⇒ ∠AMD = 90°

∴ AP ⊥ DQ.

Hence, proved that AP ⊥ DQ.

If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.

Answer

Since, AB || CD (Opposite sides of parallelogram are parallel) and BD is transversal.

From figure,

∠ABP = ∠CDQ (Alternate angles are equal)

BP = QD (As P and Q are points of trisection of the diagonal BD)

AB = CD (Opposite sides of parallelogram are equal)

Hence, △ABP ≅ △CDQ by SAS axiom.

∴ ∠APB = ∠CQD .......(By C.P.C.T.)

Multiplying both sides by -1

-∠APB = -∠CQD

Adding 180° on both sides,

⇒ 180° - ∠APB = 180° -∠CQD

⇒ ∠APQ = ∠CQP

The above angles are alternate angles and since they are equal we can say,

AP || CQ.

Hence, proved that AP || CQ.

A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD. Prove that

(i) ACBD is a rectangle.

(ii) CD is parallel to the original parallel lines.

Answer

From figure,

LM || PQ and AB is transversal.

AC, AD, BC and BD are bisectors of ∠LAB, ∠BAM, ∠PBA and ∠ABQ respectively.

So,

∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8.

(i) ∠LAB + ∠BAM = 180° (As LAM is a straight line)

$\dfrac{1}{2}$(∠LAB + ∠BAM) = $\dfrac{1}{2}$ x 180°

∠2 + ∠3 = 90° [Since, AC and AD are bisector of ∠LAB and ∠BAM]

∠CAD = 90°

∠A = 90°.

∠PBA + ∠QBA = 180° (As PBQ is a straight line)

$\dfrac{1}{2}$(∠PBA + ∠QBA) = $\dfrac{1}{2}$ x 180°

∠6 + ∠7 = 90° [Since, BC and BD are bisector of ∠PBA and ∠QBA]

∠CBD = 90°

∠B = 90°.

∠LAB + ∠ABP = 180° (LM || PQ, sum of co-interior angles is 180°)

$\dfrac{1}{2}$(∠LAB + ∠ABP) = $\dfrac{1}{2}$ x 180°

∠2 + ∠6 = 90° [Since, AC and BC are bisector of ∠LAB and ∠PBA]

In △ABC,

⇒ ∠2 + ∠6 + ∠C = 180°

⇒ 90° + ∠C = 180°

⇒ ∠C = 90°.

∠MAB + ∠ABQ = 180° (LM || PQ, sum of co-interior angles is 180°)

$\dfrac{1}{2}$(∠MAB + ∠ABQ) = $\dfrac{1}{2}$ x 180°

∠3 + ∠7 = 90° [Since, AD and BD are bisector of ∠MAB and ∠ABQ]

In △ABD,

⇒ ∠3 + ∠7 + ∠D = 180°

⇒ 90° + ∠D = 180°

⇒ ∠D = 90°.

From figure,

∠BAM = ∠ABP (Alternate angles are equal)

$\dfrac{∠\text{BAM}}{2} = \dfrac{∠\text{ABP}}{2}$

∠3 = ∠6

∠LAB = ∠ABQ (Alternate angles are equal)

$\dfrac{∠\text{LAB}}{2} = \dfrac{∠\text{ABQ}}{2}$

∠2 = ∠7

In △ABC and △ABD,

∠3 = ∠6 (Proved above)

∠2 = ∠7 (Proved above)

AB = AB (Common)

Hence, △ABC ≅ △ABD by ASA axiom.

∴ AD = BC and AC = BD (By C.P.C.T.)

Since, ∠A = ∠B = ∠C = ∠D = 90° and AD = BC, AC = BD.

Hence, proved that ACBD is a rectangle.

(ii) In △OAD,

OA = OD (Diagonals of rectangle bisect each other)

∠3 = ∠9 (Angles opposite to equal side are equal)

Since, ∠3 = 4,

∠9 = ∠4

Since, ∠9 and ∠4 are alternate angles and since they are equal we can say that,

⇒ OD || LM

⇒ CD || LM

Since, LM || PQ and CD || LM

⇒ CD || PQ.

Hence, proved that CD is parallel to the original parallel lines.

In parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2AD. Prove that

(i) BE bisects ∠B

(ii) ∠AEB = a right angle.

Answer

Parallelogram ABCD is shown in the figure below:

(i) AB = CD (Opposite sides of parallelogram are equal)

AD = BC (Opposite sides of parallelogram are equal)

From figure,

⇒ ∠1 = ∠2 (AE bisects ∠A)

⇒ ∠1 = ∠5 (Alternate angles are equal)

⇒ ∠2 = ∠5

⇒ AD = DE (As sides opposite to equal angles are equal)

Since, AB = 2AD and CD = AB

⇒ CD = 2AD

⇒ 2AD = DE + EC

⇒ 2AD = AD + EC

⇒ EC = AD.

Since, EC = AD and BC = AD

⇒ ∠6 = ∠4 (As angles opposite to equal sides are equal)

⇒ ∠6 = ∠3 (Alternate angles are equal)

∴ ∠3 = ∠4.

Since, ∠3 = ∠4, hence proved that BE bisects ∠B.

(ii) Let ∠1 = x and ∠3 = y.

∴ ∠2 = x and ∠4 = y

Since, AD || BC, sum of co-interior angles = 180

⇒ ∠A + ∠B = 180°

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ x + x + y + y = 180°

⇒ 2x + 2y = 180°

⇒ x + y = 90° ..........(1)

In △AEB,

⇒ ∠1 + ∠3 + ∠AEB = 180°

⇒ x + y + ∠AEB = 180°

⇒ 90° + ∠AEB = 180° (From i)

⇒ ∠AEB = 90°.

Hence, proved that ∠AEB = 90°.

ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2AD.

Answer

ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC at E.

To prove: AB = 2AD

Since, AE and BE are bisector of ∠A and ∠B

∠1 = ∠2 and ∠3 = ∠4

In parallelogram ABCD, we have

AB || DC

∠1 = ∠5 [Alternate angles are equal, AE is transversal]

Thus,

∠2 = ∠5 … (i)

∴ DE = AD [∵ Sides opposite to equal angles in ∆AED]

∠3 = ∠6 [Alternate angles]

∠3 = ∠4 [Since, BE is bisector of ∠B (given)]

Thus, ∠4 = ∠6 … (ii)

∴ BC = EC [∵ Sides opposite to equal angles in ∆BCE]

AD = BC [Opposite sides of || gm are equal]

AD = DE = EC

AB = DC [Opposite sides of a || gm are equal]

⇒ AB = DE + EC

⇒ AB = AD + AD

⇒ AB = 2AD

Hence, proved that AB = 2AD.

ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO = AP, prove that 3∠POB = ∠AOP.

Answer

In square ABCD, AC is a diagonal.

So, ∠CAB = 45° (As diagonals bisect vertex angle)

∠OAP = 45°

In ∆AOP,

∠OAP = 45°

AO = AP [Given]

∠AOP = ∠APO = x (let) [Angles opposite to equal sides are equal]

Now,

∠AOP + ∠APO + ∠OAP = 180° [Angles sum property of a triangle]

∠AOP + ∠AOP + 45° = 180°

2x = 180° – 45°

x = $\dfrac{135°}{2}$

∠AOB = 90° [Diagonals of a square bisect at right angles]

So, ∠AOP + ∠POB = 90°

$\dfrac{135°}{2}$ + ∠POB = 90°

∠POB = 90° – $\dfrac{135°}{2}$

= $\dfrac{45°}{2}$

3∠POB = $3 \times \dfrac{45°}{2} = \dfrac{135°}{2}$ = ∠AOP.

Hence, proved that 3∠POB = ∠AOP.

ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.

Answer

Given, AE = BF = CG = DH

Since, ABCD is a square and AB = BC = CD = AD.

So,

⇒ AB - AE = BC - BF = CD - CG = AD - DH

⇒ EB = FC = GD = AH

Now, in ∆AEH and ∆BFE

⇒ AE = BF [Given]

⇒ AH = EB [Proved]

⇒ ∠A = ∠B [Each 90°]

So, ∆AEH ≅ ∆BFE by S.A.S axiom of congruency

Then, by C.P.C.T we have

⇒ EH = EF and ∠4 = ∠2

In ∆AEH,

⇒ ∠1 + ∠4 + ∠HAE = 180°

⇒ ∠1 + ∠4 + 90° = 180°

⇒ ∠1 + ∠4 = 90°

⇒ ∠1 + ∠2 = 90° [Since, ∠4 = ∠2]

From figure,

⇒ ∠1 + ∠HEF + ∠2 = 180°

⇒ ∠HEF + 90° = 180°

⇒ ∠HEF = ∠E = 90°.

In ∆DGH and ∆CGF

⇒ DH = GC [Given]

⇒ GD = FC [Proved]

⇒ ∠D = ∠C [Each 90°]

So, ∆DGH ≅ ∆CGF by S.A.S axiom of congruency

Then, by C.P.C.T we have

GH = FG

In ∆DGH and ∆AEH

⇒ DH = AE [Given]

⇒ GD = HA [Proved]

⇒ ∠D = ∠A [Each 90°]

So, ∆DGH ≅ ∆AEH by S.A.S axiom of congruency

Then, by C.P.C.T we have

GH = HE

Thus, EF = FG = GH = HE, therefore EFGH is a Rhombus.

∵ One angle of rhombus EFGH is 90° (∠HEF = 90°),

∴ EFGH is a square.

Hence, proved that EFGH is a square.

In the figure (1) given below, ABCD and ABEF are parallelograms. Prove that

(i) CDFE is a parallelogram.

(ii) FD = EC

(iii) △AFD ≅ △BEC.

Answer

(i) DC || AB and DC = AB [∵ ABCD is a || gm] .......... (1)

FE || AB and FE = AB [∵ ABEF is a || gm] ...........(2)

∴ DC || FE and DC = FE [From (1) and (2)]

Hence, proved that CDFE is a || gm.

(ii) Since, CDEF is a || gm.

So, FD = EC (As opposite sides of a || gm are equal)

Hence, proved that FD = EC.

(iii) In ∆AFD and ∆BEC, we have

AD = BC [Opposite sides of || gm ABCD are equal]

AF = BE [Opposite sides of || gm ABEF are equal]

FD = CE [Opposite sides of || gm CDFE are equal]

Hence, ∆AFD ≅ ∆BEC by S.S.S axiom of congruency

Hence, proved that ∆AFD ≅ ∆BEC.

In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC.

Answer

From figure,

⇒ ∠FAG + ∠GAB + ∠BAD + ∠FAD = 360° [∵ At a point total angle is 360°]

⇒ ∠FAG + 90° + 90° + ∠BAD = 360°

⇒ ∠FAG = 360 – 90° – 90° – ∠BAD

⇒ ∠FAG = 180° – ∠BAD .........(i)

⇒ ∠ABC + ∠BAD = 180° [Sum of Adjacent angle in || gm is equal to 180°]

⇒ ∠ABC = 180° – ∠BAD .........(ii)

⇒ ∠FAG = ∠ABC [From (i) and (ii)]

In || gm ABCD,

AB = CD and AD = BC.

Since, ADEF is a square,

so, AD = DE = EF = FA.

So, BC = FA.

In ∆AFG and ∆ABC, we have

⇒ AF = BC [Proved above]

⇒ AG = AB (∵ AGBH is a square)

⇒ ∠FAG = ∠ABC [Proved above]

So, ∆AFG ≅ ∆ABC by S.A.S axiom of congruency

∴ FG = AC. [By C.P.C.T]

Hence, proved that FG = AC.

ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.

Answer

Rhombus ABCD is shown in the figure below:

In △ABD,

⇒ AB = AD (Sides of rhombus are equal.)

⇒ ∠B = ∠D = x (let) (∵ angles opposite to equal sides are equal)

⇒ ∠A + ∠B + ∠D = 180°

⇒ 60° + x + x = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = 60°.

∴ ABD is an equilateral triangle.

So, BD = AB = AD = a (let)

Since, diagonals of rhombus bisect each other,

OB = $\dfrac{a}{2}$

In right angled triangle AOB,

AB² = AO² + OB²

a² = AO² + $\Big(\dfrac{a}{2}\Big)^2$

AO² = a² - $\dfrac{a^2}{4} = \dfrac{3a^2}{4}$

AO = $\dfrac{\sqrt{3}a}{2}$

AC = 2AO = $\sqrt{3}a$

AC : BD = $\sqrt{3}a : a = \sqrt{3} : 1$.

Hence, AC : BD = $\sqrt{3} : 1$.