Rectilinear Figures — Exercise 13(C)

If the opposite sides of a quadrilateral are equal, the quadrilateral is :

1. rectangle

2. parallelogram

3. square

4. rhombus

Answer

If the opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.

Hence, Option 2 is the correct option.

If the opposite angles of a quadrilateral are equal, the quadrilateral is :

1. rectangle

2. parallelogram

3. square

4. rhombus

Answer

If the opposite angles of a quadrilateral are equal, the quadrilateral is a parallelogram.

Hence, Option 2 is the correct option.

If three angles of a quadrilateral are equal to 90° each, then the quadrilateral is :

1. rectangle

2. square

3. parallelogram

4. rhombus

Answer

Given,

Three angles of a quadrilateral are equal to 90° each.

We know that,

Sum of angles of a quadrilateral = 360°.

Let fourth angle be x.

∴ 3 × 90° + x = 360°

⇒ 270° + x = 360°

⇒ x = 360° - 270° = 90°.

We know that,

Each interior angle of a rectangle and square equals to 90°.

Each square is a rectangle but not each rectangle is a square.

Hence, Option 1 is the correct option.

If three angles of a quadrilateral are equal, then the quadrilateral is :

1. rectangle

2. rhombus

3. not a parallelogram

4. parallelogram

Answer

If three angles of a quadrilateral are equal, then the quadrilateral is not a parallelogram.

Hence, Option 3 is the correct option.

BEC is an equilateral triangle inside the square ABCD. The value of angle ECD is :

1. 60°

2. 30°

3. 75°

4. 45°

Answer

Given,

BEC is an equilateral triangle.

∴ ∠BCE = 60°.

From figure,

⇒ ∠ECD = ∠BCD - ∠BCE = 90° - 60° = 30°.

Hence, Option 2 is the correct option.

E is the mid-point of side AB and F is the mid point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.

Answer

We know that,

Opposite sides of parallelogram are equal.

∴ AB = CD

⇒ $\dfrac{AB}{2} = \dfrac{CD}{2}$

⇒ AE = FD.

Also,

Opposite sides of parallelogram are parallel.

∴ AB || CD

⇒ AE || FD.

∴ AE = FD and AE || FD.

Since, one pair of opposite side of quadrilateral AEFD is parallel.

Hence, proved that AEFD is a parallelogram.

The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.

Answer

Since, opposite angles of a parallelogram are equal.

∴ ∠ABC = ∠ADC = x (let)

Given,

BD bisects ∠B.

∴ ∠ABD = ∠CBD = $\dfrac{1}{2} ∠ABC = \dfrac{x}{2}$

BD bisects ∠D.

∴ ∠ADB = ∠BDC = $\dfrac{1}{2} ∠ADC = \dfrac{x}{2}$

⇒ ∠ABD = ∠ADB and ∠CBD = ∠CDB

In △ ABD,

⇒ ∠ABD = ∠ADB

∴ AB = AD (Sides opposite to equal angles are equal) ......(1)

In △ CBD,

⇒ ∠CBD = ∠CDB

∴ CD = BC (Sides opposite to equal angles are equal) .........(2)

As, ABCD is a parallelogram.

Thus, opposite sides are equal.

∴ AB = CD .........(3)

∴ AD = BC ..........(4)

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

⇒ AB = BC = CD = AD.

Since, all sides of quadrilateral ABCD are equal.

Hence, proved that ABCD is a rhombus.

The alongside figure shows a parallelogram ABCD in which AE = EF = FC. Prove that :

(i) DE is parallel to FB

(ii) DE = FB

(iii) DEBF is a parallelogram.

Answer

Join BD. Let BD intersect AC at point O.

(i) We know that,

Diagonals of parallelogram bisect each other.

∴ OB = OD and OA = OC.

From figure,

⇒ OA = OC

⇒ OA - AE = OC - FC (As, AE = FC)

⇒ OE = OF.

In quadrilateral DEBF,

⇒ OB = OD and OE = OF.

Since, diagonals of quadrilateral DEBF bisect each other,

∴ DEBF is a parallelogram.

∴ DE || FB (Opposite sides of parallelogram are parallel.)

Hence, proved that DE || FB.

(ii) We know that,

Opposite sides of parallelogram are equal.

In parallelogram DEBF,

∴ DE = FB.

Hence, proved that DE = FB.

(iii) Since, one pair of opposite sides of quadrilateral DEBF is equal and parallel,

i.e. DE || FB and DE = FB,

∴ DEBF is a parallelogram.

Hence, proved that DEBF is a parallelogram.

In the alongside figure, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B. Prove that :

(i) AQ = BP

(ii) PQ = CD

(iii) ABPQ is a parallelogram

Answer

(i) Let ∠A = 2x.

We know that,

Sum of consecutive angles in a parallelogram equals to 180°.

⇒ ∠A + ∠B = 180°

⇒ 2x + ∠B = 180°

⇒ ∠B = 180° - 2x.

⇒ ∠PAB = $\dfrac{∠A}{2} = \dfrac{2x}{2}$ = x.

⇒ ∠QBA = $\dfrac{∠B}{2} = \dfrac{180° - 2x}{2}$ = 90° - x.

In △ ABP,

⇒ ∠PAB + ∠ABP + ∠BPA = 180° (By angle sum property of triangle)

⇒ x + 180° - 2x + ∠BPA = 180°

⇒ 180° - x + ∠BPA = 180°

⇒ ∠BPA = 180° - 180° + x = x.

∴ ∠BPA = ∠PAB (Both equal to x)

∴ AB = BP (In a triangle sides opposite to equal angles are equal) ..........(1)

In △ ABQ,

⇒ ∠QBA + ∠BAQ + ∠AQB = 180° (By angle sum property of triangle)

⇒ 90° - x + 2x + ∠AQB = 180°

⇒ 90° + x + ∠AQB = 180°

⇒ ∠AQB = 180° - 90° - x = 90° - x.

∴ ∠AQB = ∠QBA (Both equal to 90° - x)

∴ AB = AQ (In a triangle sides opposite to equal angles are equal) ..........(2)

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

⇒ AQ = BP.

Hence, proved that AQ = BP.

(ii) Given,

ABCD is a parallelogram.

We know that,

Opposite sides of parallelogram are equal and parallel.

∴ AB = CD ............(1)

∴ AD || BC.

Since, AD || BC

∴ AQ || BP

Join PQ.

We know that,

AQ = BP (Proved above)

In quadrilateral ABPQ,

AQ = BP and AQ || BP.

Since, one of the pair of opposite sides of quadrilateral ABPQ are equal and parallel.

∴ ABPQ is a parallelogram.

∴ AB = PQ [Opposite sides of parallelogram are equal] .........(2)

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

PQ = CD.

Hence, proved that PQ = CD.

(iii) In quadrilateral ABPQ,

AQ || BP and AQ = BP.

∴ ABPQ is a parallelogram (Since, one of the pair of opposite sides of quadrilateral ABPQ are equal and parallel.)

Hence, proved that ABPQ is a parallelogram.

In the given figure, ABCD is a parallelogram. Prove that : AB = 2BC.

Answer

Given,

ABCD is a parallelogram.

∴ AB || CD (Opposite sides of parallelogram are parallel)

From figure,

AE is the transversal.

⇒ ∠BAE = ∠AED (Alternate angles are equal) ........(1)

⇒ ∠DAE = ∠BAE (Since, AE is the bisector of angle A) .........(2)

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

⇒ ∠AED = ∠DAE.

In △ DAE,

⇒ ∠AED = ∠DAE

⇒ AD = DE (In a triangle, sides opposite to equal angles are equal) .......(3)

From figure,

BE is the transversal.

⇒ ∠CEB = ∠EBA (Alternate angles are equal) ........(4)

⇒ ∠CBE = ∠EBA (Since, BE is the bisector of angle B) .........(5)

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

⇒ ∠CEB = ∠CBE.

In △ CBE,

⇒ ∠CEB = ∠CBE

⇒ BC = CE (In a triangle, sides opposite to equal angles are equal) .......(6)

In parallelogram ABCD,

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

⇒ AB = DE + EC

⇒ AB = AD + BC [From equation (3) and (6)]

⇒ AB = BC + BC (AD = BC, as opposite sides of parallelogram are equal)

⇒ AB = 2 BC.

Hence, proved that AB = 2 BC.

Prove that the bisectors of opposite angles of a parallelogram are parallel.

Answer

From figure,

DE and BF are bisectors of angles D and B respectively.

In parallelogram ABCD,

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

⇒ $\dfrac{∠B}{2} = \dfrac{∠D}{2}$

⇒ ∠FBC = ∠ADE.

In △ ADE and △ CBF,

⇒ ∠ADE = ∠FBC (Proved above)

⇒ AD = BC (Opposite sides of || gm ABCD are equal)

⇒ ∠DAE = ∠BCF (Opposite angles of || gm ABCD are equal)

∴ △ ADE ≅ △ CBF (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ AE = CF

Since, AE = CF and AB = CD,

∴ BE = DF

In parallelogram ABCD,

⇒ AB || CD

⇒ BE || DF

Since,

BE = DF and BE || DF

In quadrilateral BEDF, one of the pair of opposite sides are equal and parallel.

∴ BEDF is a parallelogram.

∴ DE || BF.

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

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Answer

Let ABCD be the parallelogram.

From figure,

AH, BE, CF and DG are bisectors of ∠A, ∠B, ∠C and ∠D respectively.

We know that,

Consecutive angles of parallelogram are supplementary.

Considering ∠A + ∠B = 180°,

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

⇒ $\dfrac{∠A + ∠B}{2}$ = 90°

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2}$ = 90° ............(1)

Considering ∠B + ∠C = 180°,

⇒ $\dfrac{∠B + ∠C}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠B + ∠C}{2}$ = 90°

⇒ $\dfrac{∠B}{2} + \dfrac{∠C}{2}$ = 90° ............(2)

Considering ∠C + ∠D = 180°,

⇒ $\dfrac{∠C + ∠D}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠C + ∠D}{2}$ = 90°

⇒ $\dfrac{∠C}{2} + \dfrac{∠D}{2}$ = 90° .........(3)

Considering ∠D + ∠A = 180°,

⇒ $\dfrac{∠D + ∠A}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠D + ∠A}{2}$ = 90°

⇒ $\dfrac{∠D}{2} + \dfrac{∠A}{2}$ = 90° .........(4)

In △ PAB,

By angle sum property of triangle,

⇒ ∠PAB + ∠ABP + ∠BPA = 180°

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

⇒ 90° + ∠BPA = 180° [From equation (1)]

⇒ ∠BPA = 180° - 90° = 90°.

From figure,

⇒ ∠SPQ = ∠BPA = 90° (Vertically opposite angles are equal)

In △ BSC,

By angle sum property of triangle,

⇒ ∠SBC + ∠SCB + ∠BSC = 180°

⇒ $\dfrac{∠B}{2} + \dfrac{∠C}{2}$ + ∠BSC = 180°

⇒ 90° + ∠BSC = 180° [From equation (2)]

⇒ ∠BSC = 180° - 90° = 90°.

From figure,

⇒ ∠PSR = ∠BSC = 90°.

In △ DRC,

By angle sum property of triangle,

⇒ ∠RCD + ∠CDR + ∠DRC = 180°

⇒ $\dfrac{∠C}{2} + \dfrac{∠D}{2}$ + ∠DRC = 180°

⇒ 90° + ∠DRC = 180° [From equation (3)]

⇒ ∠DRC = 180° - 90° = 90°.

From figure,

⇒ ∠SRQ = ∠DRC = 90° (Vertically opposite angles are equal)

In △ AQD,

By angle sum property of triangle,

⇒ ∠QAD + ∠ADQ + ∠DQA = 180°

⇒ $\dfrac{∠A}{2} + \dfrac{∠D}{2}$ + ∠DQA = 180°

⇒ 90° + ∠DQA = 180° [From equation (4)]

⇒ ∠DQA = 180° - 90° = 90°.

From figure,

⇒ ∠RQP = ∠DQA = 90°.

Since, all the interior angles of quadrilateral PQRS equals to 90°.

∴ PQRS is a rectangle.

Hence, proved that bisectors of interior angles of a parallelogram form a rectangle.

Prove that the bisectors of the interior angles of a rectangle form a square.

Answer

In rectangle,

All the interior angles equal to 90°. So, bisectors divide interior angles into two 45° angles.

In △ BSC,

By angle sum property of triangle,

⇒ ∠SBC + ∠SCB + ∠BSC = 180°

⇒ $\dfrac{∠B}{2} + \dfrac{∠C}{2}$ + ∠BSC = 180°

⇒ 45° + 45° + ∠BSC = 180°

⇒ ∠BSC = 180° - 90° = 90°.

Since,

⇒ ∠SBC = ∠SCB

∴ BS = SC (Sides opposite to equal angles are equal) .....(1)

From figure,

⇒ ∠PSR = ∠BSC = 90°.

In △ APB and △ DRC,

⇒ ∠PAB = ∠RDC (Both equal to 45°)

⇒ ∠PBA = ∠RCD (Both equal to 45°)

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

∴ △ APB ≅ △ DRC (By A.S.A. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BP = CR .........(2)

Subtracting equation (2) from (1), we get :

⇒ BS - BP = SC - CR

⇒ PS = SR .............(3)

In △ DRC,

By angle sum property of triangle,

⇒ ∠RCD + ∠CDR + ∠DRC = 180°

⇒ $\dfrac{∠C}{2} + \dfrac{∠D}{2}$ + ∠DRC = 180°

⇒ 45° + 45° + ∠DRC = 180°

⇒ ∠DRC = 180° - 90° = 90°.

From figure,

⇒ ∠SRQ = ∠DRC = 90° (Vertically opposite angles are equal)

In △ PAB,

By angle sum property of triangle,

⇒ ∠PAB + ∠ABP + ∠BPA = 180°

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

⇒ 45° + 45° + ∠BPA = 180°

⇒ ∠BPA = 180° - 90° = 90°.

From figure,

⇒ ∠SPQ = ∠BPA = 90° (Vertically opposite angles are equal)

In △ AQD,

By angle sum property of triangle,

⇒ ∠DAQ + ∠QDA + ∠AQD = 180°

⇒ $\dfrac{∠A}{2} + \dfrac{∠D}{2}$ + ∠AQD = 180°

⇒ 45° + 45° + ∠AQD = 180°

⇒ ∠AQD = 180° - 90° = 90°.

Since,

⇒ ∠DAQ = ∠QDA

∴ AQ = QD (Sides opposite to equal angles are equal) .....(4)

From figure,

⇒ ∠PQR = ∠AQD = 90°.

Since, △ APB ≅ △ DRC

∴ AP = DR .........(5)

Subtracting equation (5) from (4), we get :

⇒ AQ - AP = QD - DR

⇒ PQ = QR ...........(6)

In △ BSC and △ AQD,

⇒ ∠B = ∠A (Both equal to 45°)

⇒ ∠C = ∠D (Both equal to 45°)

⇒ ∠S = ∠Q (Both equal to 90°)

∴ △ BSC ≅ △ AQD (By A.A.A. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BS = AQ .........(7)

In △ AQD,

⇒ ∠A = ∠D (Both equal to 45°)

⇒ DQ = AQ (Sides opposite to equal angles are equal) .........(8)

From equation (7) and (8), we get :

⇒ BS = DQ .........(9)

In △ CDR,

⇒ ∠C = ∠D (Both equal to 45°)

⇒ DR = CR (Sides opposite to equal angles are equal) .........(10)

From equation (2) and (10), we get :

⇒ BP = DR .....(11)

Subtracting equation (11) from (9), we get :

⇒ BS - BP = DQ - DR

⇒ PS = QR .........(12)

From equation (3), (6) and (12), we get :

⇒ PQ = QR = RS = PS.

Since, all sides of quadrilateral PQRS are equal and each interior angle equals to 90°.

∴ PQRS is a square.

Hence, proved that bisectors of the interior angles of a rectangle form a square.

In parallelogram ABCD, the bisectors of angle A meets DC at P and AB = 2AD.

Prove that :

(i) BP bisects angle B.

(ii) Angle APB = 90°.

Answer

(i) Let AD = x.

Given,

AB = 2AD = 2x.

Given,

AP is the bisector of angle A,

∴ ∠1 = ∠2 .............(1)

From figure,

⇒ ∠2 = ∠5 [Alternate angles are equal] ......(2)

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

⇒ ∠1 = ∠5

In △ ADP,

⇒ ∠1 = ∠5

⇒ DP = AD = x (Sides opposite to equal angles are equal)

From figure,

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

⇒ CD = 2x

⇒ DP + PC = 2x

⇒ x + PC = 2x

⇒ PC = 2x - x = x.

Also,

⇒ BC = AD = x (Opposite sides of parallelogram are equal)

In △ BCP,

⇒ BC = PC (Both equal to x)

⇒ ∠6 = ∠4 (Angles opposite to equal sides are equal) ......(3)

⇒ ∠6 = ∠3 (Alternate angles are equal) ..........(4)

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

⇒ ∠3 = ∠4.

∴ BP is the bisector of angle B.

Hence, proved that BP bisects angle B.

(ii) We know that,

Consecutive angles of parallelogram are supplementary.

Considering ∠A + ∠B = 180°,

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

⇒ $\dfrac{∠A + ∠B}{2}$ = 90°

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2}$ = 90° ............(1)

In △ PAB,

By angle sum property of triangle,

⇒ ∠PAB + ∠ABP + ∠APB = 180°

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

⇒ 90° + ∠APB = 180° [From equation (1)]

⇒ ∠APB = 180° - 90° = 90°.

Hence, proved that ∠APB = 90°.

Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.

Answer

Join BD. Let BD intersect AC at point O.

In parallelogram ABCD,

Diagonals of || gm bisect each other.

⇒ OA = OC .........(1)

⇒ OB = OD

Given,

⇒ AM = CN .........(2)

Subtracting equation (2) from (1), we get :

⇒ OA - AM = OC - CN

⇒ OM = ON.

In quadrilateral BMDN,

⇒ OM = ON and OB = OD.

∴ Diagonals of quadrilateral BMDN bisect each other.

∴ BMDN is a parallelogram.

Hence, proved that BMDN is a parallelogram.

In the following figure, ABCD is a parallelogram. Prove that :

(i) AP bisects angle A

(ii) BP bisects angle B

(iii) ∠DAP + ∠CBP = ∠APB

Answer

(i) In △ ADP,

⇒ AD = DP (Given)

∴ ∠APD = ∠PAD (Angles opposite to equal sides are equal) .......(1)

In parallelogram ABCD,

AB || DC and AP is the transversal.

∴ ∠APD = ∠PAB (Alternate angles are equal) ........(2)

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

⇒ ∠PAD = ∠PAB.

∴ AP bisects angle A.

Hence, proved that AP bisects angle A.

(ii) In || gm ABCD,

⇒ BC = AD (Opposite sides of || gm are equal)

∴ BC = PC

∴ ∠BPC = ∠PBC (Angles opposite to equal sides are equal) .......(3)

In parallelogram ABCD,

AB || DC and BP is the transversal.

∴ ∠BPC = ∠PBA (Alternate angles are equal) ........(4)

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

⇒ ∠PBC = ∠PBA.

∴ BP bisects angle B.

Hence, proved that BP bisects angle B.

(iii) We know that,

Consecutive angles of a || gm are supplementary.

∴ ∠A + ∠B = 180°,

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

⇒ $\dfrac{∠A + ∠B}{2}$ = 90°

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

⇒ ∠PAB + ∠PBA = 90° .........(1)

In △ APB,

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ 90° + ∠APB = 180°

⇒ ∠APB = 180° - 90°

⇒ ∠APB = 90° ........(2)

From figure,

⇒ ∠PAB = ∠DAP (As, AP bisects ∠A)

⇒ ∠PBA = ∠CBP (As, BP bisects ∠B)

Substituting value of ∠PAB and ∠PBA in equation (1), we get :

⇒ ∠DAP + ∠CBP = 90° .........(3)

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

⇒ ∠DAP + ∠CBP = ∠APB.

Hence, proved that ∠DAP + ∠CBP = ∠APB.

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

Let AP and DQ intersect at point O.

In △ DAQ and △ ABP,

⇒ ∠DAQ = ∠ABP (Interior angle of square equal to 90°)

⇒ DQ = AP (Given)

⇒ AD = AB (Each side of square equal in length)

∴ △ DAQ ≅ △ ABP (By R.H.S. congruence rule)

We know that,

Corresponding parts of congruent triangle are equal.

∴ ∠3 = ∠1 ........(1)

From figure,

⇒ ∠1 + ∠4 = 90°

Substituting value of ∠1 from equation (1) in above equation, we get :

⇒ ∠3 + ∠4 = 90°

In triangle AOD,

By angle sum property of triangle,

⇒ ∠ODA + ∠OAD + ∠AOD = 180°

⇒ ∠3 + ∠4 + ∠AOD = 180°

⇒ 90° + ∠AOD = 180°

⇒ ∠AOD = 180° - 90° = 90°.

∴ AP ⊥ DQ.

Hence, proved that AP and DQ are perpendicular to each other.

In a quadrilateral ABCD, AB = AD and CB = CD. Prove that :

(i) AC bisects angle BAD.

(ii) AC is perpendicular bisector of BD.

Answer

(i) In △ ABC and △ ADC,

⇒ AB = AD (Given)

⇒ BC = CD (Given)

⇒ AC = AC (Common side)

∴ △ ABC ≅ △ ADC (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ ∠BAC = ∠DAC

∴ AC bisects ∠BAD.

Hence, proved that AC bisects angle BAD.

(ii) Since, AC bisects ∠BAD

∴ ∠BAO = ∠DAO

In △ AOB and △ AOD,

⇒ AB = AD (Given)

⇒ AO = AO (Common side)

⇒ ∠BAO = ∠DAO (Proved above)

∴ △ AOB ≅ △ AOD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ ∠BOA = ∠DOA ........(1)

From figure,

⇒ ∠BOA + ∠DOA = 180° (Linear pair)

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

⇒ 2∠BOA = 180°

⇒ ∠BOA = $\dfrac{180°}{2}$ = 90°.

∴ AC is perpendicular bisector of BD.

Hence, proved that AC is perpendicular bisector of BD.