Quadrilaterals

Let ABCD be a parallelogram with equal diagonals.

From figure,

In ∆ ABC and ∆ DCB,

⇒ AB = DC (Opposite sides of a parallelogram are equal)

⇒ BC = BC (Common side)

⇒ AC = DB (Diagonals of parallelogram are equal)

∴ ∆ ABC ≅ ∆ DCB (By S.S.S. Congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ ∠ABC = ∠DCB (By C.P.C.T.) ......(1)

We know that,

Sum of co-interior angles equal to 180°.

⇒ ∠ABC + ∠DCB = 180° (AB || CD)

⇒ ∠ABC + ∠ABC = 180° [From equation (1)]

⇒ 2∠ABC = 180°

⇒ ∠ABC = $\dfrac{180°}{2}$

⇒ ∠ABC = ∠B = 90°

⇒ ∠DCB = ∠C = 90°

Since,

Opposite angles of parallelogram are equal.

∴ ∠D = ∠B = 90° and ∠A = ∠C = 90°.

∴ ∠B = ∠D = ∠C = ∠A = 90°.

Since, opposite sides of ABCD are equal in length and each interior angle equals to 90°.

Hence, proved that if the diagonals of a parallelogram are equal, then it is a rectangle.

Let ABCD be the square.

In Δ ABC and Δ BAD,

⇒ AB = AB (Common side)

⇒ BC = AD (Each side of a square is equal in length)

⇒ ∠ABC = ∠BAD = 90° (Each interior angle in a square is a right angle)

∴ Δ ABC ≅ Δ BAD (By S.A.S. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

∴ AC = BD (By C.P.C.T.)

In Δ OAD and Δ OCB,

⇒ AD = CB (Each side of a square is equal in length)

⇒ ∠OAD = ∠OCB (Alternate angles are equal)

⇒ ∠ODA = ∠OBC (Alternate angles are equal)

∴ Δ OAD ≅ Δ OCB (By A.S.A. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ OA = OC (By C.P.C.T.) ......(1)

⇒ OB = OD (By C.P.C.T.) .....(2)

In Δ OBA and Δ ODA,

⇒ OB = OD ....[From equation (2)]

⇒ OA = OA (Common side)

⇒ BA = DA (Each side of a square is equal in length)

∴ Δ OBA ≅ Δ ODA (By S.S.S. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ ∠AOB = ∠AOD = x (let) (By C.P.C.T.)

⇒ ∠AOB + ∠AOD = 180° (Linear pair)

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

⇒ ∠AOB = 90° and ∠AOD = 90°.

Thus, AC and BD bisect each other at right angles.

Hence, proved that the diagonals of a square are equal and bisects each other at right angles.

(i) We know that,

ABCD is a parallelogram.

So, opposite sides are parallel and equal.

⇒ ∠DAC = ∠BCA (Alternate interior angles are equal) ......(1)

⇒ ∠BAC = ∠DCA (Alternate interior angles are equal) .......(2)

Given,

⇒ AC bisects ∠A

⇒ ∠DAC = ∠BAC .......(3)

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

⇒ ∠DCA = ∠BAC = ∠DAC = ∠BCA .........(4)

∴ ∠DCA = ∠BCA

∴ AC bisects ∠C.

Hence, proved that AC bisects ∠C.

(ii) In Δ ABC,

⇒ ∠BAC = ∠BCA (Proved above)

⇒ BC = AB (Side opposite to equal angles are equal) .....(5)

⇒ DA = BC and AB = CD (Opposite sides of a parallelogram are equal) ....(6)

From equations (5) and (6), we get :

⇒ AB = BC = CD = DA.

Since, all sides of quadrilateral ABCD are equal.

Hence, proved that ABCD is a rhombus.

Rectangle ABCD is shown in the figure below:

(i) Given :

ABCD is a rectangle and AC bisects ∠A and ∠C.

⇒ ∠DAC = ∠CAB ....(1)

⇒ ∠DCA = ∠BCA .....(2)

We know that,

Opposite sides of a rectangle are parallel and equal.

From figure,

AD || BC and AC is transversal,

⇒ ∠DAC = ∠BCA ..........(3) (Alternate interior angles are equal)

From equations (1) and (3), we get :

⇒ ∠CAB = ∠BCA ......(4)

In △ ABC,

⇒ ∠CAB = ∠BCA

We know that,

Sides opposite to equal angles are equal.

⇒ BC = AB .....(5)

We know that,

Opposite sides of a rectangle are equal.

⇒ BC = AD .........(6)

⇒ AB = DC .........(7)

From equation (5), (6) and (7), we get :

⇒ AB = BC = CD = AD.

Since,

ABCD is a rectangle and all the sides are equal. Hence, ABCD is a square.

Hence, proved that ABCD is a square.

(ii) Join BD.

In Δ BCD,

⇒ BC = CD (Sides of a square are equal to each other)

⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal) ..... (8)

⇒ ∠CDB = ∠ABD (Alternate interior angles are equal) ..... (9)

From equations (8) and (9), we get :

⇒ ∠CBD = ∠ABD

∴ BD bisects ∠B.

From figure,

⇒ ∠CBD = ∠ADB (Alternate interior angles are equal) .....(10)

From equations (8) and (10), we get :

⇒ ∠ADB = ∠CDB

∴ BD bisects ∠D.

Hence, proved that diagonal BD bisects ∠B as well as ∠D.

Given :

ABCD is a parallelogram.

From figure,

AD || BC and BD is transversal

⇒ ∠ADP = ∠CBQ (Alternate angles are equal)

(i) In Δ APD and Δ CQB,

⇒ AD = CB (Opposite sides of parallelogram ABCD are equal)

⇒ DP = BQ (Given)

⇒ ∠ADP = ∠CBQ (Proved above)

∴ Δ APD ≅ Δ CQB (By S.A.S. congruence rule)

Hence, proved that Δ APD ≅ Δ CQB.

(ii) In Δ APD ≅ Δ CQB,

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AP = CQ (By C.P.C.T.) ........(1)

Hence, proved that AP = CQ.

(iii) In Δ AQB and Δ CPD,

⇒ AB = CD (Opposite sides of parallelogram ABCD are equal)

⇒ ∠ABQ = ∠CDP (Alternate interior angles are equal)

⇒ BQ = DP (Given)

⇒ Δ AQB ≅ Δ CPD (By S.A.S. congruence rule)

Hence, proved that Δ AQB ≅ Δ CPD.

(iv) Since Δ AQB ≅ Δ CPD,

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AQ = CP (C.P.C.T.) ............(2)

Hence, proved that AQ = CP.

(v) From equation (1) and (2), we get :

⇒ AQ = CP and AP = CQ

Since both pairs of opposite sides in APCQ are equal,

Hence, proved that APCQ is a parallelogram.

Given :

ABCD is a parallelogram with AP ⊥ BD and CQ ⊥ BD.

From figure,

AB || DC and BD is transversal.

(i) In Δ APB and Δ CQD,

⇒ ∠APB = ∠CQD (Both equal to 90°)

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

⇒ ∠ABP = ∠CDQ (Alternate interior angles are equal)

∴ Δ APB ≅ Δ CQD (By A.A.S. congruence rule)

Hence, proved that Δ APB ≅ Δ CQD.

(ii) Since,

Δ APB ≅ Δ CQD

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AP = CQ (By C.P.C.T.)

Hence, proved that AP = CQ.

Steps of construction :

1. Extend AB.

2. Draw a line through C parallel to DA intersecting AB produced at E.

Given :

ABCD is a trapezium, AB || CD and AD = BC

By construction,

AD || CE

and

AE || DC.

In AECD, both pair of opposite sides are parallel, hence, AECD is a parallelogram

(i) From figure,

AD = CE (Opposite sides of parallelogram are equal)

As,

AD = BC

∴ BC = CE

We know that,

Angles opposite to equal sides in a triangle are also equal.

⇒ ∠CEB = ∠CBE .........(1)

Consider, parallel lines AD and CE where AE is the transversal.

We know that,

Sum of co-interior angles = 180°

⇒ ∠BAD + ∠CEB = 180°

⇒ ∠BAD + ∠CBE = 180° [From equation (1)] ..... (2)

From figure,

⇒ ∠ABC + ∠CBE = 180° (Linear pairs) ....... (3)

From equations (2) and (3), we get :

⇒ ∠BAD + ∠CBE = ∠ABC + ∠CBE

⇒ ∠BAD = ∠ABC.

∴ ∠A = ∠B

Hence, proved that ∠A = ∠B.

(ii) From figure,

AB || CD

We know that,

Sum of co-interior angles equal to 180°.

⇒ ∠A + ∠D = 180° .........(4)

⇒ ∠C + ∠B = 180° .........(5)

From equation (4) and (5), we get :

⇒ ∠A + ∠D = ∠C + ∠B

Since, ∠A = ∠B

We can say that,

⇒ ∠D = ∠C

Hence, proved that ∠C = ∠D.

(iii) Join AC and BD.

In ∆ ABC and ∆ BAD,

⇒ AB = BA (Common side)

⇒ BC = AD (Given)

⇒ ∠B = ∠A (above)

∴ ∆ ABC ≅ ∆ BAD (By S.A.S. congruence rule)

Hence, proved that Δ ABC ≅ Δ BAD.

(iv) Since ∆ ABC ≅ ∆ BAD,

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AC = BD (By C.P.C.T.)

Hence, proved that diagonal AC = diagonal BD.

Given :

ABCD is a quadrilateral, where P, Q, R and S are the mid points of the sides AB, BC, CD and DA.

Mid-point theorem : The line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.

(i) In △ ADC,

S and R are the mid-points of side AD and CD respectively.

By mid-point theorem,

⇒ SR || AC .......(1)

⇒ SR = $\dfrac{1}{2}AC$ ......(2)

Hence, proved that SR || AC and SR = $\dfrac{1}{2}AC$.

(ii) In Δ ABC, P and Q are mid-points of sides AB and BC.

By using the mid-point theorem,

⇒ PQ || AC .........(3)

⇒ PQ = $\dfrac{1}{2}AC$ ......(4)

From equations (3) and (4), we get :

⇒ PQ = SR

Hence, proved that PQ = SR.

(iii) From equation (1) and (3), we get :

⇒ PQ || AC || SR

⇒ PQ || SR

Also,

⇒ PQ = SR (Proved above)

We know that,

If one pair of opposite sides are equal and parallel, then the figure is parallelogram.

Hence, proved that PQRS is a parallelogram.

In Δ ABC,

P and Q are mid points of AB and BC

By mid-point theorem,

PQ || AC and PQ = $\dfrac{1}{2}AC$ .....(1)

In Δ ADC,

S and R are mid points of AD and DC.

By mid-point theorem,

⇒ SR || AC and SR = $\dfrac{1}{2}AC$ .....(2)

From equations (1) and (2), we get :

⇒ SR = PQ and SR || PQ.

Since, one of the opposite pairs of PQRS are parallel and equal.

∴ PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

In Δ BAC,

P and Q are mid points of AB and BC.

By mid-point theorem,

PQ || AC and PQ = $\dfrac{1}{2}AC$

∴ MQ || ON

In Δ BCD,

Q and R are mid points of BC and CD.

By mid-point theorem,

QR || BD and QR = $\dfrac{1}{2}BD$

∴ QN || OM

Since, opposite sides of quadrilateral OMQN are parallel.

∴ OMQN is a parallelogram

⇒ ∠MQN = ∠NOM (Opposite angles of parallelogram are equal)

We know that,

Diagonals of rhombus intersect at 90°

∴ ∠NOM = 90°

⇒ ∠PQR = ∠NOM = 90°

So, ∠PQR = 90°

In PQRS,

One pair of opposite side is parallel and equal and one of its interior angle is 90°.

Hence, proved that PQRS is a rectangle.

We know that,

The diagonals of a rectangle are equal.

⇒ BD = AC = x (let)

In △ ABC,

P and Q are the mid-points of AB and BC respectively.

By mid-point theorem,

⇒ PQ || AC and PQ = $\dfrac{1}{2}AC = \dfrac{x}{2}$ ....(1)

In △ ADC,

S and R are the mid-points of AD and CD respectively.

SR || AC and SR = $\dfrac{1}{2}AC = \dfrac{x}{2}$ .....(2)

From equation (1) and (2), we get :

PQ || SR and PQ = SR

In quadrilateral PQRS, one pair of opposite sides are equal and parallel to each other.

∴ PQRS is a parallelogram.

In △ BCD, Q and R are the mid-points of side BC and CD respectively.

By mid-point theorem,

⇒ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{x}{2}$ ....(3)

In △ BAD, P and S are the mid-points of side AB and AD respectively.

By mid-point theorem,

PS || BD and PS = $\dfrac{1}{2}BD = \dfrac{x}{2}$ .......(4)

From equations (1), (2), (3), (4), we get :

PQ = QR = SR = PS

Hence, proved that the quadrilateral PQRS is a rhombus.

Given :

ABCD is a trapezium, where AB || DC, E is the mid-point of AD and EF || AB.

By converse of mid-point theorem,

A line drawn through the mid-point of any side of a triangle and parallel to another side bisects the third side.

Let EF intersect BD at point G.

In trapezium ABCD,

EF || AB and E is the mid-point of AD.

By converse of mid-point theorem,

G is the mid-point of DB

As, EF || AB and AB || CD,

EF || CD (Two lines parallel to the same line are parallel to each other)

In Δ BCD,

GF || CD and G is the mid-point of line BD.

By using the converse of mid-point theorem, F is the mid-point of BC.

Hence, proved that F is the mid-point of BC.

Given :

ABCD is a parallelogram, where E and F are the mid-points of sides AB and CD.

We know that,

Opposite sides of a parallelogram are equal and parallel.

∴ AB || CD and AB = CD

∴ AE || FC

Since, AB = CD

Dividing both sides of equation by 2, we get :

⇒ $\dfrac{1}{2}AB$ = $\dfrac{1}{2}CD$

⇒ AE = FC

∴ AEFC is a parallelogram.

∴ AF || CE

From figure,

⇒ PF || QC

⇒ EQ || AP

In △ DQC,

F is the mid-point of DC and FP || CQ.

By converse of mid-point theorem, we get :

P is the mid-point of DQ.

⇒ DP = PQ .....(1)

In △ APB,

E is the mid-point of AB and EQ || AP.

By converse of mid-point theorem, we get :

Q is the mid-point of BP.

⇒ PQ = QB ......(2)

From equations (1) and (2) we get :

⇒ PQ = QB = DP.

∴ AF and EC trisect BD.

Hence, proved that AF and EC trisect the diagonal BD.

(i) In Δ ABC,

It is given that M is the mid-point of AB and MD || BC.

By converse of mid-point theorem, we get :

Thus, D is the mid-point of AC.

Hence, proved that D is the mid-point of AC.

(ii) As DM || CB and AC is a transversal,

We know that,

Sum of co-interior angles = 180°.

⇒ ∠MDC + ∠DCB = 180°

⇒ ∠MDC + 90° = 180°

⇒ ∠MDC = 180° - 90°

⇒ ∠MDC = 90°

Hence, proved that MD ⊥ AC.

(iii) Join MC,

In Δ AMD and Δ CMD,

⇒ AD = CD (D is the mid-point of side AC)

⇒ ∠ADM = ∠CDM (Since, MD ⊥ AC)

⇒ DM = DM (Common side)

∴ Δ AMD ≅ Δ CMD (By S.A.S. congruence rule)

We know that,

Corresponding parts of congruent triangles are equal.

⇒ AM = CM (By C.P.C.T.)

We know that,

M is the mid-point of AB.

∴ AM = $\dfrac{1}{2}AB$

⇒ CM = AM = $\dfrac{1}{2}AB$.

Hence, proved that CM = MA = $\dfrac{1}{2}AB$.