(i) ∠XYZ = 135°    ... [Given]

WXYZ is a parallelogram.

∠XWZ = ∠XYZ

∴ ∠XWZ = 135°           ... (1)

(ii) ∠YZW + ∠XYZ = 180° ... [Adjacent angles of a parallelogram are supplementary]

∴ ∠YZW + 135° = 180°         ... [From (1)]

∴ ∠YZW = 180° - 135°

∴ ∠YZW = 45°

(iii) l(OY) = 5 cm         ... [Given]

l(OY) –  \(\frac{1}{2}\)l(WY)            ... [Diagonals of a parallelogram bisect each other]

∴ l(WY) = 2 × l(OY)

∴ l(WY) = 2 × 5

∴ l(WY) = 10 cm

Answer is : ∠XWZ = 135°, ∠YZW = 45°, l(WY) = 10 cm

□ ABCD is a parallelogram.

∴ ∠A + ∠B = 180°        ... (Adjacent angles of parallelogram are supplementary)

∴ (3x + 12)° + (2x - 32)° = 180°

∴ 3x + 12 + 2x – 32 = 180

∴ 5x - 20 = 180

∴ 5x = 180 + 20

∴ 5x = 200

∴ x = 200/5

∴ x = 40

∠C = ∠A                      ... (Opposite angles of a parallelogram)

∴ ∠C = (3x + 12)°

∴ ∠C = [3(40) + 12]°

∴ ∠C = (120 + 12)°

∴ ∠C = 132°

∠D = ∠B                      ... (Opposite angles of a parallelogram)

∴ ∠D = (2x - 32)°

∴ ∠D = [2(40) - 32]°

∴ ∠D = (80 - 32)°

∴ ∠D = 48°

Answer is : The value of x is 40, ∠C = 132°, ∠D = 48°.

Let □ ABCD be the given parallelogram.

Let AB = x cm

According to the given condition,

BC = (x + 25) cm

CD = AB = x cm                   ... (Opposite sides of a parallelogram are congruent)

AD = BC = (x+25) cm          ... (Opposite sides of a parallelogram are congruent)

The perimeter of parallelogram ABCD = 150 cm                 ... (Given)

∴ AB + BC + CD + AD = 150

∴ x + (x + 25) + x + (x + 25) = 150

∴ 4x + 50 = 150

∴ 4x = 150 - 50

∴ 4x = 100

∴ x = 100/4

∴ x = 25  ∴ CD = AB = x = 25 cm.

AD = BC = x + 25 = 25 + 25 = 50 cm

Answer is : The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 em and 50 cm.

Let □ ABCD be the parallelogram.

The ratio of measures of two adjacent angles of a parallelogram is 1 : 2.

Let the common multiple be x.

∴ ∠A = x° and ∠B = 2x°

∠A + ∠B = 180°

∴ X + 2x = 180       ... [Adjacent angles of a parallelogram are supplementary]

∴ 3x = 180

x = 180/3 = 60

∠A = x° = 60°

∠B = 2x° = 2×60° = 120°

∠A = ∠C = 60°

∠B = ∠D = 120°                   ... [Opposite angles of a parallelogram]

Answer is : The angles of parallelogram are 60°, 120°, 60° and 120°.

AO = 5, BO = 12, AB = 13             ... (Given)

AB² = 13² = 169

AO² + OB² = 52 + 122

∴ AO² + OB² = 25 + 144

∴ AO² + OB² = 169

∴ AO² + OB² = AB²

By converse of Pythagoras theorem,

∠AOB = 90°                 ... (1)

In parallelogram ABCD, diagonal AC ⊥ diagonal BD    ... [From (1)]

Also, diagonals AC and BD bisect each other. ... (Property of a parallelogram)

􀀍 ABCD is a rhombus ... (A parallelogram is a rhombus if its diagonals are perpendicular bisector of each other.)

􀀍 PQRS is a parallelogram             .... [Given]

∴ ∠R = ZP                    ... [Opposite angles of a parallelogram]

∴ ∠R = 110°                 ... (i)

􀀍ABCR is a parallelogram.             ... [Given]

∴ ∠A + ∠R = 180°        ... [Adjacent angles of a parallelogram are supplementary]

∴ ∠A + 110° = 180°     ... [From (i)]

∴ ∠A = 180° - 110°

∴ ∠A = 70°

∴ ∠C = ∠A = 70°

∴ ∠B = ∠R = 110° ... [Opposite angles of a parallelogram]

Answer is : ∠R = 110°, ∠B = 110°, ∠A = 70°, ∠C =70°.

seg AB || seg DC          ... (Opposite sides of a parallelogram)

∴ seg AE || seg DC       ... (A-B-E)

and line BC is the transversal,

∴ ∠EBC ≅ ∠DCB            ... (Alternate angles)

i.e. ∠EBF ≅ ∠DCF         ... (B-F-C) ... (1)

seg AB ≅ seg CD          ... (Opposite sides of a parallelogram) ... (2)

also seg AB ≅ seg BE    ... (Given) ... (3)

∴ seg BE ≅ seg CD       [From (2) and (3)] ... (4)

seg AE || seg DC and line DE is the transversal.

∴ ∠BED ≅ ∠EDC           ... (Alternate angles)

i.e. ∠BEF ≅ ∠FDC         ... (5)

In Δ EBF and Δ DCF,   

seg BE ≅ seg CD          ... [From (4)]

∠EBF ≅ ∠DCF      ... [From (1)]

∠BEF ≅ ∠FDC      ... [From (5)]

∴ Δ EBF ≅ Δ DCF         ... (ASA test of congruence)

∴ seg BF ≅ seg CF        ... (c.s.c.t)

∴ line ED bisects seg BC at point F.

Given : 􀀍 ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively.

To Prove : 􀀍 APCQ is a parallelogram.

Proof:

AP = \(\frac{1}{2}\)AB           ... (i) [P is the midpoint of side AB]

QC = \(\frac{1}{2}\)DC           ... (ii) [Q is the midpoint of side CD]

􀀍 ABCD is a parallelogram.   ... [Given]

∴ AB = DC  ... [Opposite sides of a parallelogram]

\(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC ... [Multiplying both sides by \(\frac{1}{2}\) ]

∴ AP = QC ... (iii) [From (i) and (ii)]

Also, AB || DC ... [Opposite angles of a parallelogram]

i.e. AP || QC        ... (iv) [A-P-B, D-Q-C]

From (iii) and (iv),

AP = QC

AP || QC

A quadrilateral is a parallelogram if its opposite sides is parallel and congruent.

∴ 􀀍 APCQ is a parallelogram.

Let 􀀍 ABCD be a rectangle.

All angles are right angles: ∠A = ∠B = ∠C = ∠D = 90^∘

Thus, ∠A =∠C and ∠B =∠D.

The opposite angles test states: If both pairs of opposite angles are equal, the quadrilateral is a parallelogram.

Therefore, every rectangle is a parallelogram by this test.

Proof : Let the median from vertex D intersect side EF at M.

G is the centroid          ... (Given)

Centroid divides each median in the ratio 2 : 1

∴ DG : GM = 2 : 1

∴ \(\frac{DG}{GM}=\frac{2}{1}\)

∴ DG = 2GM       ... (1)

DG = GH             ... (Given)

∴ DG = GM + MH         ... (G-M-H)

∴ 2GM = GM + MH       ... [From (1)]

∴ 2GM - GM = MH

∴ GM = MH         ... (2)

In 􀀍 GEHF,

seg GM ≅ seg MH         ... [From (2)]

seg EM ≅ seg MF          ... (M is the midpoint of seg EF)

A quadrilateral is a parallelogram, if its diagonals bisect each other.

∴ 􀀍 GEHF is a parallelogram.

Proof :

□ ABCD is a parallelogram    ... (Given)

∠ADC + ∠BCD = 180°  ... (Adjacent angles of a parallelogram are supplementary)

Multiplying each term by \(\frac{1}{2}\), we get,

\(\frac{1}{2}\)∠ADC + \(\frac{1}{2}\)∠BCD = \(\frac{1}{2}\) × 180°

But,   \(\frac{1}{2}\)∠ADC = ∠PDC    ... (Ray DP bisects ∠ADC) ... (2)

And ∠BCD = ∠PCD     ... (Ray CP bisects ∠BCD) ... (3)

∴ ∠PDC + ∠PCD = 90°          ... [From (1), (2) and (3)] ... (4)

In Δ PDC,

∠PDC + ∠PCD + ∠DPC = 180° ... (The sum of the measures of all angles of a triangle is 180°)

∴ 90° + ∠DPC = 180°          ... [From (4)]

∴ ∠DPC=180° - 90°             

∴ ∠DPC = 90°                     

i.e. ∠SPQ = 90°           ... (D-S-P, P-Q-C) ... (5)

Similarly, we can prove SRQ = 90° ... (6)

also ∠ASD = 90° and ∠BQC = 90°          ... (7)

∠PSR = ∠ASD     ... (Vertically opposite angles)

∴ ∠PSR = 90°              (From 7) ... (8)

Similarly, ∠PQR = 90°           ... (9)

In □ PQRS,

∠SPQ = ∠SRQ = ∠PSR = ∠PQR = 90° ... (From 5, 6, 8 and 9)

∴ □ PQRS is a rectangle ... (By definition)

Proof : AP = CR                   ... (Given) ... (1)

□ ABCD is a parallelogram    ... (Given)

∴ AB = CD                            ... (Opposite sides of a parallelogram)

∴ AP + PB = CR + RD            ... (A-P-B, C-R-D) ... (2)

∴ PB = RD                            ... [From (1) and (2)] ... (3)

∠ABC ≅ ∠ADC ... (Opposite angles of a parallelogram)

i.e. ∠PBQ ≅ ∠RDS                 ... (A-P-B, B-Q-C, C-R-D and A-S-D) ... (4)

In Δ PBQ and Δ RDS,

seg PB ≅ seg RD                  ... [From (3)]

PBQ ≅ ZRDS                        ... [From (4)]

seg BQ ≅ seg SD                  ... (Given)

∴ Δ PBQ ≅ Δ RDS                  ... (SAS test of congruence)

∴ seg PQ ≅ seg RS                ... (c.s.c.t) ... (5)

Similarly, by proving Δ PAS ≅ Δ RCQ, we get,           ... (6)

seg PS ≅ seg RQ

In □ PQRS,

seg PQ ≅ seg RS          ... [From (5)]

seg PS ≅ seg RQ          ... [From (6)]

A quadrilateral is a parallelogram if its opposite sides are congruent.

∴ □ PQRS is a parallelogram.

□ ABCD is a rectangle ... [Given]

∴ BD = AC          ... [Diagonals of a rectangle are congruent]

∴ BD = 8 cm       ... [From (i)]

BO = \(\frac{1}{2}\)BD          ... [Diagonals of a rectangle bisect each other]

∴ BO = \(\frac{1}{2}\) × 8

∴ BO = 4 cm

side AD || side BC and seg AC is their transversal. ... [Opposite sides of a rectangle are parallel]

∴ ∠ACB = ∠CAD           ... [Alternate angles]

∠ACB = 35°                 ...[ ∵ ∠CAD = 35°]

∴ BO = 4 cm, ∠ACB = 35°

PQ= 7.5 cm         ... (Given)

PQ = QR             ... (Sides of a rhombus)

∴ QR = 7.5 cm

∠QPS = 75°        ... (Given)

∠QPS + ∠PQR = 180°  ... (Adjacent angles of a rhombus are supplementary)

∴ 75° + ∠PQR = 180°

∴ ∠PQR = 180° - 75°

∴ ∠PQR = 105°

∠SRQ = ∠QPS     ... (Opposite angles of a rhombus)

∴ ∠SRQ = 75°

Answer is : QR = 7.5 cm, ∠PQR = 105°, ∠SRQ = 75°.

□ IJKL is a square.

Diagonals of a square are perpendicular to each other

∴ ∠IMJ = 90°

Diagonals of a square bisects opposite angles.

∴ ∠JIK = \(\frac{1}{2}\)∠JIL

∴ ∠JIK = \(\frac{1}{2}\) x 90°

∴ ∠JIK = 45°

Also, ∠LJK = \(\frac{1}{2}\)∠IJK

∴ ∠LJK = \(\frac{1}{2}\) × 90°

∴ ∠LJK = 45°

Answer is : ∠IMJ = 90°, ∠JIK = 45° and ∠LJK = 45°.

Let □ ABCD be given rhombus.

AC = 20 cm, BD = 21 cm.

Let diagonals AC and BD intersect at point O.

AO = \(\frac{1}{2}\)AC and BO = \(\frac{1}{2}\)BD              ... (Diagonals of a rhombus bisect each other)

∴ AO = \(\frac{1}{2}\) × 20  ∴ AO = 10 cm

and BO = \(\frac{1}{2}\) x 21 = 10.5 cm

Ιη ΔΑΟΒ,

∴ ∠AOB = 90°             ... (Diagonals of a rhombus are perpendicular to each other)

By Pythagoras theorem,

AB² = AO² + OB²

∴ AB² = 10² + 10.5²

∴ AB² = 100 + 110.25

∴ AB² = 210.25

∴ AB = 210.25

∴ AB = 14.5 cm

Side of the rhombus = AB = 14.5 cm

Perimeter of the rhombus = 4 x side =4 × 14.5 = 58cm

Answer is : Side of the rhombus is 14.5 cm Perimeter of the rhombus is 58 cm.

False.

• In a rhombus all sides are congruent. Whereas in a parallelogram only opposite sides are congruent. Hence every parallelogram is not a rhombus.

False.

In a rectangle all angles are of measure 90°. Whereas in a rhombus it is not necessary that all angles are of measure 90°.

(iii) Every rectangle is a parallelogram.

True.

• A quadrilateral is a parallelogram, if its opposite sides are parallel. In a rectangle, opposite sides are parallel. Hence every rectangle is a parallelogram.

True.

• A quadrilateral is a rectangle, if all its angles are right angles. A square has all of its angles right angles. Hence every square is a rectangle.

True.

A quadrilateral is a rhombus, if all of its sides are congruent. In a  square, all sides are congruent. Hence, every square is a rhombus.

False.

• A quadrilateral is a rectangle, if all of its angles are right angles. In a parallelogram it is not necessary that all angles are right angles. Hence every parallelogram is not a rectangle.

seg LK || seg IJ and line LI is the transversal,

∠L + ∠I = 180°           …..(Interior angles)

∴ ∠L + 108° = 180°

∴ ∠L = 180° - 108°

∴ ∠L = 72°

seg LK || seg IJ and line KJ is the transversal,

∠K + ∠J = 180°           …..(Interior angles)

∴ 53° + ∠J = 180°

∴ ∠J = 180° - 53°

∴ ∠J = 127°

Answer is : ∠L = 72° and ∠J = 127°.

Draw seg BM ⊥ side AD such that A-M-D and seg CN ⊥ side AD such that A-N-D.

seg BC || seg AD          ... (Given)

∴ BM = CN          ... (Distance between two parallel lines is constant) ... (1)

In A BMA and ACND,

∠BMA = ∠CND = 90°   ... (Construction)

Hypotenuse BA ≅ Hypotenuse CD           ... (Given)

side BM side CN           ... [From (1)]

∴ ∠BMA = ∠CND          ... (Hypotenuse side test of congruence)

∴ ∠BAM ≅ ∠CDN          ... (c.a.c.t.)

i.e. ∠BAD ≅ ∠CDA        ... (A-M-D, A-N-D)

∠BAD = 72°                ... (Given)

∠CDA = 72° i.e. ∠D = 72°

side BC || side AD        ... (Given)

and line BA is the transversal,

∠BAD + ∠ABC = 180° …..(Interior angles)

∴ 72° + ∠ABC = 180°

∴ ∠ABC = 180° - 72°

∴ ∠ABC = 108° i.e. ∠B = 108°

Answer is : ∠D = 72° and ∠B = 108°.

Proof :

Draw seg BM ⊥ side AD such that A-M-D and seg CN ⊥ side AD such that A-N-D.

side BC || side AD                 ... (Given)

∴ BM = CN          ... (Distance between two parallel lines is constant) ... (1)

In ABMA and A CND,

∠BMA = ∠CND = 90°            ….(Construction)

Hypotenuse BA ≅ Hypotenuse CD   ... (Given)

side BM ≅ side CN                     ... [From (1)]

∴ Δ BMA ≅ Δ CND ... (By hypotenuse side test of congruence)

∴ ∠BAM ≅ ∠CDN                   ... (c.a.c.t.)

i.e. ∠BAD ≅ ∠CDA                ... (A-M-D, A-N-D) ... (2)

side BC || ≅ side AD              ... (Given)

and line AB is the transversal,

∴ ∠ABC + ∠BAD = 180°                ... (Interior angles) ... (3)

side BC || side AD                 ... (Given)

and line CD is the transversal,

∴ ∠DCB + ∠CDA = 180°        ... (Interior angles) ... (4)

∴ ∠ABC + ∠BAD = ∠DCB + ∠CDA ... [From (3) and (4)]

∴ ∠ABC + ∠CDA = ∠DCB + ∠CDA  ... [From (2)]

∴ ∠ABC = ∠DCB

∴ ∠ABC ≅ ∠DCB.

AB = 5 cm, AC = 9 cm and BC = 11 cm   ... (Given)

In Δ ABC, X and Y are the midpoints of sides AB and BC respectively. ... (Given)

∴ XY = \(\frac{1}{2}\)AC        ... (Midpoint theorem)

∴ XY = \(\frac{1}{2}\) x 9

∴ XY = 4.5 cm

In Δ ABC, Y and Z are the midpoints of sides BC and AC respectively. ... (Given)

∴ YZ = \(\frac{1}{2}\)AB                 ... (Midpoint theorem)

∴ YZ = \(\frac{1}{2}\) x 5

∴ YZ = 2.5 cm

In Δ ABC, X and Z are the midpoints of sides AB and AC respectively         ... (Given)

∴ XZ = \(\frac{1}{2}\)BC                ... (Midpoint theorem)

∴ XZ = \(\frac{1}{2}\) x 11

∴ XZ = 5.5 cm

Answer is : XY = 4.5 cm, YZ = 2.5 cm, XZ = 5.5 cm.

Proof :

(i) 􀀍LMNR is a rectangle.

∴ seg LM || seg RN               ... (Opposite sides of a rectangle)

i.e. seg LM || seg RQ             ... (R -N - Q) ... (1)

seg RQ || seg SP                  ... (Opposite sides of a rectangle) ... (2)

From (1) and (2),

seg LM || seg SP                   ... (3)

In Δ RSP, M is the midpoint of seg PR      ... (Given)

seg LM || seg SP                   ... [From (3)]

∴ L is the midpoint of side SR         ... (By converse of midpoint theorem) ... (4)

∴ SL = LR

(ii) Diagonals of a rectangle are congruent.

∴ SQ = PR and LN = MR

Now, MR = \(\frac{1}{2}\)PR  ... (M is the midpoint of seg PR) ... (7)

∴ LN = \(\frac{1}{2}\) PR                 ... [From (6) and (7)] ... (8)

∴ LN = \(\frac{1}{2}\) SQ.               ... [From (5) and (8)]

Proof :

In Δ ABC, F and E are the midpoints of sides AB and AC respectively.          ... (Given)

∴ FE = \(\frac{1}{2}\)BC                 ... (Midpoint theorem) ... (1)

In Δ ABC, D and E are the midpoints of sides BC and AC respectively.

∴ DE = \(\frac{1}{2}\)AB                ... (Midpoint theorem) ... (2)

In Δ ABC, D and F are the midpoints of sides BC and AB respectively.

DF = \(\frac{1}{2}\)AC                   ... (Midpoint theorem) ... (3)

Δ ABC is a equilateral.

∴ BC = AB = AC           ... (Sides of an equilateral triangle)

Multiplying each term by \(\frac{1}{2}\), we get,

\(\frac{1}{2}\)BC = \(\frac{1}{2}\)AB = \(\frac{1}{2}\)AC                ... (4)

∴ FE = DE = DF           ... [From (1), (2), (3) and (4)]

∴ Δ DEF is an equilateral triangle.   ... (By definition)

Proof:

Draw seg DN || seg QM such that P-M-N and M-N-R.

In Δ PDN,

Point T is the mid-point of seg PD.           ... (Given)

seg TM || seg DN         ... (Construction)

∴ Point M is the mid-point of seg PN. ... (Converse of mid-point theorem) [P-M-N]

∴ PM = MN

In Δ QMR,

Point D is the mid-point of seg QR.          ... (Given)

seg DN || seg QM.                ... (Construction)

∴ Point N is the mid-point of seg MR. ... (Converse of mid-point theorem)[M-N-R]

∴ RN = MN                  ... (ii)

PM = MN = RN             ….[From (i) and (ii)] ... (iii)

Now,

PR = PM + MN + RN

∴ PR = PM + PM + PM          ……[From (iii)]

∴ PR = 3PM

∴ \(\frac{PR}{PM}\) = \(\frac{3}{1}\)

Hence proved.

(D) rhombus

• When all the pairs of the adjacent sides of a quadrilateral are congruent then it is a rhombus.

(C) 48 cm

Explanation :

In Δ ABC, we apply the Pythagoras theorem,

AB² + BC² = AC²

x² + x² = AC²               (□ABCD is a square.)

2x² = AC²

2x² = (12\(\sqrt{2}\) )²

2x² = 144(2)

x² = 144

x = 12 cm

∴ x = AB = BC = CD = AD = 12 cm

Perimeter of □ABCD = 4 x 12 = 48 cm

(D) 40 °

A rhombus is also a parallelogram so, the opposite angles will be congruent.

Thus, (2x)° =(3x - 40)°

3x - 2x = 40°

x = 40°

Let □ ABCD be a given rectangle.

AB = 7 cm, BC = 24 cm

In Δ ABC,

∠ABC = 90°        ... (Angle of a rectangle)

By Pythagoras theorem,

AC² = AB² + BC²

∴ AC² = 7² + 24²

∴ AC² = 49 + 576

AC² = 625

∴ AC = \(\sqrt{625}\)

∴ AC = 25 cm

∴ The length of the diagonal is 25 cm.

Let □ ABCD be a given square.

AC = 13 cm

Let the side of the square be x cm.

In Δ ABC,

∠ABC = 90°

By Pythagoras theorem,

AC² = AB² + BC²

∴ 13² = x² + x²

∴ 2x² = 169

∴ x² = 169/2

∴ x = \(\sqrt{169/2}\) = \(\frac{13}{\sqrt{2}}\) = \(\frac{13}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\) = 6.5\(\sqrt{2}\)

Side of a square is 6.5\(\sqrt{2}\)

Let □ ABCD be a given parallelogram.

AB : BC = 3 : 4

Let the common multiple of the given ratio be x.

Then AB = 3x cm,

BC = 4x

CD = AB = 3x cm         ... (Opposite sides of a parallelogram)

AD = BC = 4x cm         ... (Opposite sides of a parallelogram)

Perimeter of □ ABCD = 112 cm

∴ AB + BC + CD + AD = 112

∴ 3x + 4x + 3x + 4x = 112

∴ 14x = 112

∴ 4x = 112/14

∴ x = 8.

∴ CD =AB = 3x = 3 × 8 = 24 cm

AD = BC = 4x = 4 x 8 = 32 cm.

∴ The sides of the parallelogram are of lengths 24 cm, 32 cm, 24 cm and 32 cm.

Let the diagonals PR and QS intersect at point O.

PR = 20 cm, QS = 48 cm

The diagonals of a rhombus bisect each other at right angles       ... (1)

∴ OP = \(\frac{1}{2}\)PR = \(\frac{1}{2}\) × 20 cm

∴ OP = 10 cm              ... (2)

OQ = \(\frac{1}{2}\) QS = \(\frac{1}{2}\) × 48 cm

∴ OQ = 24 cm              ... (3)

∠POQ = 90°                ... [From (1)]

In right angled Δ POQ, by Pythagoras theorem,

PQ² = OP² + OQ²

= (10)² + (24)2           ... [From (2) and (3)]

= 100 + 576 = 676

∴ PQ² = (26)²

∴ PQ = 26 cm

∴ The length of side PQ is 26 cm.

□ PQRS is rectangle

Diagonals of a rectangle are equal.

.. ∴ PR = QS.

Multiplying both the sides by \(\frac{1}{2}\)

\(\frac{1}{2}\)PR = \(\frac{1}{2}\)QS

Now, PM = \(\frac{1}{2}\)PR            ... (2) ... (Diagonals of rectangle bisect each other)

MS = \(\frac{1}{2}\)QS                   ... (Diagonals of rectangle bisect each other)

∴ PM = MS                   ... [From (1), (2) and (3)]

∴ ∠MPS ≅ ∠MSP          ... (Isosceles triangle theorem)

Let ∠MPS = ∠MSP = x  ... (4)

∠PMS = ∠QMR             ... (Vertically opposite angles)

∠QMR = 50°                ... (Given)

∴ ∠PMS = 50°              ... (5)

In Δ MPS,

∠PMS + ∠MPS + ∠MSP = 180° ... (The sum of the measures of all angles of a triangle is 180°)

∴ 50° + x + x = 180°            ... [From (5) and (4)]

∴ 2x = 180°- 50°

∴ 2x = 130°

∴ x = 130°/2

∴ x = 65°  

∴ ∠MPS = x = 65°

∴ The measure of MPS is 65°.

In □ ABQP,

seg AB || seg PQ          ... (Given)

seg AB seg PQ             ... (Given)

A quadrilateral is a parallelogram, if a pair of opposite sides is parallel and congruent.

□ ABQP is a parallelogram

∴ seg AP || seg BQ       ... (1)

and seg AP ≅ seg BQ    ... (2) ... (Opposite sides of a parallelogram are parallel and congruent)

In □ ACRP,

seg AC || seg PR, seg AC ≅ seg PR          ... (Given)

A quadrilateral is a parallelogram, if a pair of opposite sides is parallel and congruent.

□ ACRP is a parallelogram

∴ seg AP || seg CR       ... (3)

seg AP ≅ seg CR          ... (4) ... (Opposite sides of a parallelogram are parallel and congruent)

In □ BQRC,

seg BQ || seg CR          ... [From (1) and (3)]

seg BQ ≅ seg CR          ... [From (2) and (4)]

A quadrilateral is a parallelogram, if a pair of opposite sides is parallel and congruent.

□ BQRC is a parallelogram

seg BC || seg QR,

seg BC ≅ seg QR ... (Opposite sides of a parallelogram are parallel and congruent)

Construction : Draw seg AQ and extend it to intersect side DC produce at R.

Proof : line AB | line DC and line BC is the transversal                ... (Given)

∴ ∠ABC = ∠RCB           ... (Alternate angles)

i.e. ∠ABQ = ∠RCQ       ... (B-Q-C) ... (1)

In Δ ABQ and Δ RCQ,

∠ABQ ≅ ∠RCQ

seg BQ ≅ seg CQ         ... (Q is the midpoint of seg BC)

∠BQA ≅ ∠CQR             ... (Vertically opposite angles)

∴ Δ ABQ ≅ Δ RCQ        ... (ASA test of congruence)

∴ seg AB ≅ seg RC       ... (c.s.c.t.) ... (2)

∴ seg AQ ≅ seg QR      ... (c.s.c.t.) ... (3)

In Δ ADR,

P is the midpoint of seg AD            ... (Given)

Q is the midpoint of seg AR            ... (From 3)

∴ seg PQ || side DR              ... (Midpoint theorem)

∴ seg PQ || side DC              ... (D-C-R) ... (4)

side DC || side AB                 ... (Given) ... (5)

∴ seg PQ || side AB               ... [From (4) and (5)]

also, PQ = \(\frac{1}{2}\) DR                    ... (Midpoint theorem)

∴ PQ = \(\frac{1}{2}\)(DC + CR)              ... (D-C-R)

∴ PQ = \(\frac{1}{2}\)(DC + AB)              ... [From (2)]

i.e. PQ = \(\frac{1}{2}\)(AB + DC).

Proof : Draw seg DM and extend it to intersect side AB at point T.  ... (A-T-B) ... (Given)

side DC || side AB and line AC is the transversal

∴ ∠DCA ≅ ∠BAC                   ... (Alternate angles)

i.e. ∠DCM ≅ ∠TAM                ... (A-M-C and A-T-B) ... (1)

In Δ DCM and Δ TAM,

∠DCM ≅ ∠TAM

seg MC ≅ seg MA                   ... (M is the midpoint of side AC)

∠DMC ≅ ∠TMA                      ... (Vertically opposite angles)

∴ Δ DCM ≅ Δ TAM                 ... [From (1)] ... (ASA test of congruence)