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

Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.

To Prove: Quadrilateral ABCD is a square.

Proof: In ∆OAD and ∆OCB,

OA = OC | Given

OD = OB | Given

∠AOD = ∠COB

| Vertically Opposite Angles

∴ ∆OAD ≅ ∆OCB

| SAS Congruence Rule

∴ AD = CB | C.P.C.T.

∠ODA = ∠OBC | C.P.C.T.

∴ ∠BDA = ∠DBC

∴ AD || BC

Now, ∵ AD = CB and AD || CB

∴ Quadrilateral ABCD is a || gm.

In ∆AOB and ∆AOD,

AO = AO | Common

OB = OD | Given

∠AOB = ∠AOD

| Each = 90° (Given)

∴ ∆AOB ≅ ∆AOD

| SAS Congruence Rule

∴ AB = AD

Now, ∵ ABCD is a parallelogram and

∴ AB = AD

∴ ABCD is a rhombus.

Again, in ∆ABC and ∆BAD,

AC = BD | Given

BC = AD

| ∵ ABCD is a rhombus

AB = BA | Common

∴ ∆ABC ≅ ∆BAD

| SSS Congruence Rule

∴ ∆ABC = ∆BAD | C.P.C.T.

AD || BC

| Opp. sides of || gm ABCD and transversal AB intersects them.

∴ ∠ABC + ∠BAD = 180°

| Sum of consecutive interior angles on the same side of a transversal is 180°

∴ ∠ABC = ∠BAD = 90°

Similarly, ∠BCD = ∠ADC = 90°

∴ ABCD is a square.

811 Views

Show that the diagonals of a square are equal and bisect each other at right angles.

Given: ABCD is a square.

To Prove: (i) AC = BD

(ii) AC and BD bisect each other at right angles.

Proof: (i) In ∆ABC and ∆BAD,

AB = BA | Common

BC = AD Opp. sides of square ABCD

∠ABC = ∠BAD | Each = 90°

(∵ ABCD is a square)

∴ ∆ABC ≅ ∆BAD

| SAS Congruence Rule

∴ AC = BD | C.P.C.T

(ii) In ∆OAD and ∆OCB,

AD = CB

| Opp. sides of square ABCD

∠OAD = ∠OCB

| ∵ AD || BC and transversal AC intersects them

∠ODA = ∠OBC

| ∵ AD || BC and transversal BD intersects them

∴ ∆OAD ≅ ∆OCB

| ASA Congruence Rule

∴ OA = OC ...(1)

Similarly, we can prove that

OB = OD ...(2)

In view of (1) and (2),

AC and BD bisect each other.

Again, in ∆OBA and ∆ODA,

OB = OD | From (2) above

BA = DA

| Opp. sides of square ABCD

OA = OA | Common

∴ ∆OBA ≅ ∆ODA

| SSS Congruence Rule

∴ ∠AOB = ∠AOD | C.P.C.T.

But ∠AOB + ∠AOD = 180°

| Linear Pair Axiom

∴ ∠AOB = ∠AOD = 90°

∴ AC and BD bisect each other at right angles.

923 Views

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Given: ABCD is a quadrilateral whose diagonals AC and BD intersect each other at right angles at O.

To Prove: Quadrilateral ABCD is a rhombus.

Proof: In ∆AOB and ∆AOD,

AO = AO | Common

OB = OD | Given

∠AOB = ∠AOD | Each = 90°

∴ ∆AOB ≅ ∆AOD

| SSS Congruence Rule

∴ AB = AD ...(1) | C.P.C.T.

Similarly, we can prove that

AB = BC ...(2)

BC = CD ...(3)

CD = AD ...(4)

In view of (1), (2), (3) and (4), we obtain

AB = BC = CD = DA

∴ Quadrilateral ABCD is a rhombus.

1597 Views

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Given: In parallelogram ABCD, AC = BD.

To Prove: ||gm ABCD is a rectangle.

Proof: In ∆ACB and ∆BDA,

AC = BD | Given

AB = BA | Common

BC = AD

| Opposite sides of || gm ABCD

∴ ∆ACB ≅ ∆BDA

| SSS Congruence Rule

∴ ∠ABC = ∠BAD ...(1) C.P.C.T.

Again, ∵ AD || BC

| Opp. sides of || gm ABCD and transversal AB intersects them.

∴ ∠BAD + ∠ABC = 180° ...(2)

| Sum of consecutive interior angles on the same side of a transversal is 180°

From (1) and (2),

∠BAD = ∠ABC = 90°

∴ ∠A = 90°

∴ || gm ABCD is a rectangle.

2336 Views

Let ABCD be a quadrilateral in which

∠A : ∠B : ∠C : ∠D = 3 : 5 : 9 : 13

Sum of the ratios = 3 + 5 + 9+ 13 = 30

Also, ∠A + ∠B + ∠C + ∠D = 360°

Sum of all the angles of a quadrilateral is 360°

1580 Views