Pythagoras Theorem — Test Yourself

Angle AOB is:

1. 60°

2. 90°

3. 45°

4. none of these

Answer

5, 12, and 13 are pythagorean triplets.

As,

13² = 12² + 5²

So, side BO = 13 units is the hypotenuse.

We know that,

Angle opposite to hypotenuse is a right-angle triangle.

So, ∠OAB = 90°

According to angle sum property,

⇒ ∠OAB + ∠ABO + ∠AOB = 180°

⇒ 90° + ∠ABO + ∠AOB = 180°

⇒ ∠ABO + ∠AOB = 180° - 90°

⇒ ∠ABO + ∠AOB = 90°

Thus, ∠AOB, ∠ABO cannot be defined individually.

Hence, option 4 is the correct option.

Ranbeer runs 10 km due North and 24 km due West. The distance between his two position is:

1. 34 km

2. 17 km

3. 26 km

4. none of these

Answer

Starting from point A, Ranbeer runs 10 km due north and reached at point B then 24 km due west, ending at point C.

According to Pythagoras Theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 10² + 24²

⇒ AC² = 100 + 576

⇒ AC² = 676

⇒ AC = $\sqrt{676}$

⇒ AC = 26 km.

Hence, option 3 is the correct option.

Angle AOB is:

1. 60°

2. 90°

3. 45°

4. none of these

Answer

Since, AO = BO = x

Squaring all the sides we get :

AB² = $(x\sqrt{2})^2 = 2x^2$

OB = x²

OA = x²

In triangle AOB,

⇒ AO² + BO² = x² + x² = 2x² = AB².

Thus, AOB is a right angle triangle with AB as hypotenuse.

Since, AB is hypotenuse, then ∠AOB (angle opposite to hypotenuse) = 90°

Hence, option 2 is the correct option.

The sides of a rectangle are 12 cm and 16 cm. The length of its diagonal is:

1. 28 cm

2. 4 cm

3. $\sqrt{12^2 + 16^2}$ cm

4. $\sqrt{16^2 - 12^2}$ cm

Answer

ABCD is a rectangle such that AB = 16 cm and BC = 12 cm.

Join AC. AC is a diagonal of the rectangle ABCD.

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 16² + 12²

⇒ AC = $\sqrt{16^2 + 12^2}$ cm

Hence, option 3 is the correct option.

Statement 1: ABCD is a rhombus, its diagonal AC = 16 cm and diagonal BD = 12 cm, perimeter of rhombus = 64 cm.

Statement 2: OA = 8 cm, OB = 6 cm. Then, AB = 10 cm

And, perimeter of rhombus = 40 cm

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, ABCD is a rhombus, diagonal AC = 16 cm and diagonal BD = 12 cm.

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

∴ ∠AOB = 90°

Each diagonal is divided into two equal segments.

∴ AC = 16 cm ⇒ AO = OC = $\dfrac{16}{2}$ = 8 cm

∴ BD = 12 cm ⇒ BO = OD = $\dfrac{12}{2}$ = 6 cm

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In Δ AOB,

⇒ AB² = BO² + OA²

⇒ AB² = 6² + 8²

⇒ AB² = 36 + 64

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

According to formula, perimeter of rhombus = 4 x length of side

= 4 x 10

= 40 cm.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Statement 1: Area of given triangle ABC = 6 x 5 cm².

Statement 2: Area of given triangle ABC = $\dfrac{1}{2}$ x 6 x 4 cm².

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Draw a perpendicular bisector, AD.

The perpendicular bisector of the base also passes through the midpoint of the base, effectively dividing it into two equal segments.

BD = DC = $\dfrac{6}{2}$ = 3 cm

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In Δ ABD,

⇒ AB² = AD² + BD²

⇒ 5² = AD² + 3²

⇒ 25 = AD² + 9

⇒ AD² = 25 - 9

⇒ AD² = 16

⇒ AD = $\sqrt{16}$

⇒ AD = 4 cm.

Using formula, area of triangle = $\dfrac{1}{2}$ x base x height

Here, base, BC = 6 cm and height, AD = 4 cm

⇒ Area of triangle = $\dfrac{1}{2}$ x 6 x 4 cm²

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Assertion (A): Angle BOC = 90°.

Reason (R): OC² = 3² + 4² = 25

OB² = 6² + 8² = 100

OC² + OB² = 125 = BC²

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, OD = 3 cm and DC = 4 cm

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In Δ ODC,

⇒ OC² = OD² + DC²

⇒ OC² = 3² + 4²

⇒ OC² = 9 + 16

⇒ OC² = 25

⇒ OC = $\sqrt{25}$

⇒ OC = 5 cm

Similarly, it it given that OA = 6 cm and AB = 8 cm

In Δ OAB,

⇒ OB² = OA² + AB²

⇒ OB² = 6² + 8²

⇒ OB² = 36 + 64

⇒ OB² = 100

⇒ OB = $\sqrt{100}$

⇒ OB = 10 cm

Squaring all sides of triangle BOC,

⇒ OB² = 10² = 100

⇒ OC² = 5² = 25

⇒ BC² = $(5\sqrt{5})^2$ = 125

Since,

⇒ BC² = OC² + OB²

Since, sides of triangle BOC, satisfy pythagoras theorem. So, BOC is a right angle triangle with BC as hypotenuse.

∴ ∠BOC = 90°.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): x = $5\sqrt{2}$

Reason (R): AC² = 8² + 6² = x² + x²

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

In Δ ABC,

Since angle ABC = 90°.

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

⇒ AC² = AB² + BC²

⇒ AC² = 8² + 6² ........................(1)

In Δ ADC,

Since angle ADC = 90°.

Using pythagoras theorem,

⇒ AC² = AD² + DC²

⇒ AC² = x² + x² .....................(2)

From equation (1) and (2),

⇒ 8² + 6² = x² + x²

⇒ 64 + 36 = 2x²

⇒ 2x² = 100

⇒ x² = $\dfrac{100}{2}$

⇒ x² = 50

⇒ x = $\sqrt{50}$

⇒ x = 5 $\sqrt{2}$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

In the given figure, AB // CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.

Answer

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ BD² = AD² + AB²

⇒ 25² = AD² + 7²

⇒ 625 = AD² + 49

⇒ AD² = 625 - 49

⇒ AD² = 576

⇒ AD = $\sqrt{576}$ = 24 cm.

Draw BE perpendicular to CD.

From figure,

ABED is a rectangle. Since, opposite sides of rectangle are equal.

∴ ED = AB = 7 cm and BE = AD = 24 cm.

⇒ CE = CD - ED = 17 - 7 = 10 cm.

In right-angled triangle BEC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ BC² = BE² + CE²

⇒ BC² = 24² + 10²

⇒ BC² = 576 + 100

⇒ BC² = 676

⇒ BC = $\sqrt{676}$ = 26 cm.

Hence, BC = 26 cm.

In the given figure, ∠B = 90°, XY // BC, AB = 12 cm, AY = 8 cm and AX : XB = 1 : 2 = AY : YC. Find the lengths of AC and BC.

Answer

Given,

AX : XB = 1 : 2

Let AX = x and XB = 2x.

From figure,

⇒ AX + XB = AB

⇒ x + 2x = 12

⇒ 3x = 12

⇒ x = $\dfrac{12}{3}$ = 4 cm.

⇒ AX = x = 4 cm and XB = 2x = 2(4) = 8 cm.

Given,

AY : YC = 1 : 2

$\Rightarrow \dfrac{AY}{YC} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{8}{YC} = \dfrac{1}{2} \\[1em] \Rightarrow YC = 8 \times 2 = 16.$

From figure,

⇒ AC = AY + YC = 8 + 16 = 24 cm.

In right-angled triangle ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AC² = AB² + BC²

⇒ 24² = 12² + BC²

⇒ 576 = 144 + BC²

⇒ BC² = 576 - 144

⇒ BC² = 432

⇒ BC = $\sqrt{432} = 12\sqrt{3}$ = 20.78 cm.

Hence, AC = 24 cm and BC = 20.78 cm.

In △ ABC, ∠B = 90°. Find the sides of the triangle, if :

(i) AB = (x - 3) cm, BC = (x + 4) cm and AC = (x + 6) cm

(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm.

Answer

(i) In right-angled triangle ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AC² = AB² + BC²

⇒ (x + 6)² = (x - 3)² + (x + 4)²

⇒ x² + 6² + 12x = x² + 9 - 6x + x² + 16 + 8x

⇒ x² + 36 + 12x = x² + 9 - 6x + x² + 16 + 8x

⇒ x² + 36 + 12x = 2x² + 25 + 2x

⇒ 2x² - x² + 2x - 12x + 25 - 36 = 0

⇒ x² - 10x - 11 = 0

⇒ x² - 11x + x - 11 = 0

⇒ x(x - 11) + 1(x - 11) = 0

⇒ (x + 1)(x - 11) = 0

⇒ x + 1 = 0 or x - 11 = 0

⇒ x = -1 or x = 11.

Since, side cannot be negative.

∴ x = 11.

AB = x - 3 = 11 - 3 = 8 cm,

BC = x + 4 = 11 + 4 = 15 cm,

AC = x + 6 = 11 + 6 = 17 cm.

Hence, AB = 8 cm, BC = 15 cm and AC = 17 cm.

(ii) In right-angled triangle ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AC² = AB² + BC²

⇒ (4x + 5)² = x² + (4x + 4)²

⇒ (4x)² + 5² + 2 × (4x) × 5 = x² + (4x)² + 4² + 2 × (4x) × 4

⇒ 16x² + 25 + 40x = x² + 16x² + 16 + 32x

⇒ 16x² + 25 + 40x = 17x² + 16 + 32x

⇒ 17x² - 16x² + 16 - 25 + 32x - 40x = 0

⇒ x² - 8x - 9 = 0

⇒ x² - 9x + x - 9 = 0

⇒ x(x - 9) + 1(x - 9) = 0

⇒ (x + 1)(x - 9) = 0

⇒ x + 1 = 0 or x - 9 = 0

⇒ x = -1 or x = 9.

Since, side cannot be negative.

∴ x = 9.

AB = x = 9 cm,

BC = 4x + 4 = 4(9) + 4 = 36 + 4 = 40 cm,

AC = 4x + 5 = 4(9) + 5 = 35 + 5 = 41 cm.

Hence, AB = 9 cm, BC = 40 cm and AC = 41 cm.

If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, find the other diagonal.

Answer

Let ABCD be the rhombus and the diagonals intersect at point O.

Let diagonal AC = 16 cm.

We know that,

Diagonals of rhombus bisect each other at right angles.

∴ AO = OC = $\dfrac{AC}{2} = \dfrac{16}{2}$ = 8 cm and BO = OD = x cm (let).

In right angle triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AB² = AO² + OB²

⇒ 10² = 8² + x²

⇒ 100 = 64 + x²

⇒ x² = 100 - 64

⇒ x² = 36

⇒ x = $\sqrt{36}$ = 6 cm.

From figure,

⇒ BD = BO + OD = 6 + 6 = 12 cm.

Hence, length of other diagonal = 12 cm.

In the given figure, diagonals AC and BD intersect at right angle. Show that :

AB² + CD² = AD² + BC²

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle AOB,

⇒ AB² = OB² + OA² ..........(1)

In right angle triangle COD,

⇒ CD² = OD² + OC² ..........(2)

In right angle triangle AOD,

⇒ AD² = OD² + OA² ..........(3)

In right angle triangle BOC,

⇒ BC² = OB² + OC² ..........(4)

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

⇒ AB² + CD² = OB² + OA² + OD² + OC²

⇒ AB² + CD² = (OA² + OD²) + (OC² + OB²)

Substituting value from (3) and (4) in above equation, we get :

⇒ AB² + CD² = AD² + BC².

Hence, proved that AB² + CD² = AD² + BC².

Diagonals of rhombus ABCD intersect each other at point O. Prove that :

OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle AOD,

By pythagoras theorem,

⇒ AD² = OA² + OD²

⇒ OA² = AD² - OD² ..........(1)

In right angle triangle COD,

By pythagoras theorem,

⇒ CD² = OC² + OD²

⇒ OC² = CD² - OD² ..........(2)

To prove :

OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$

Solving L.H.S. of the above equation, we get :

⇒ OA² + OC² = AD² - OD² + CD² - OD²

⇒ OA² + OC² = AD² + CD² - 2OD² .........(3)

Since, all sides of a rhombus are equal and diagonals of rhombus bisect each other,

∴ CD = AD and OD = $\dfrac{BD}{2}$

Substituting value of CD and OD in equation (3), we get :

⇒ OA² + OC² = AD² + AD² - $2\Big(\dfrac{BD}{2}\Big)^2$

⇒ OA² + OC² = 2AD² - $2 \times \dfrac{BD^2}{4}$

⇒ OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$ = R.H.S.

Hence, proved that OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$.

In the figure, AB = BC and AD is perpendicular to CD. Prove that :

AC² = 2.BC.DC.

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ AD² = AB² - BD² .........(1)

In right angle triangle ACD,

By pythagoras theorem,

⇒ AC² = AD² + DC²

= (AB² - BD²) + (DB + BC)² [From equation (1)]

= AB² - BD² + DB² + BC² + 2.DB.BC

= AB² + BC² + 2.DB.BC

= BC² + BC² + 2.DB.BC [As, AB = BC]

= 2BC² + 2.DB.BC

= 2BC(BC + DB)

= 2BC.DC

Hence, proved that AC² = 2BC.DC

In an isosceles triangle ABC; AB = AC and D is a point on BC produced. Prove that :

AD² = AC² + BD.CD

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

Draw AE perpendicular to BC.

In right angle triangle AED,

By pythagoras theorem,

⇒ AD² = AE² + ED²

⇒ AD² = AE² + (EC + CD)² .........(1)

In right angle triangle AEC,

By pythagoras theorem,

⇒ AC² = AE² + EC²

⇒ AE² = AC² - EC² .........(2)

Substituting value of AE² from equation (2) in (1), we get :

⇒ AD² = AC² - EC² + (EC + CD)²

⇒ AD² = AC² - EC² + EC² + CD² + 2.EC.CD

⇒ AD² = AC² + CD(CD + 2EC) ..........(3)

Since, ABC is an isosceles triangle.

We know that,

In an isosceles triangle altitude from the vertex bisects the base.

∴ E is the mid-point of BC.

⇒ EC = $\dfrac{1}{2}$ BC

⇒ BC = 2EC.

From figure,

⇒ BD = BC + CD

⇒ BD = 2EC + CD

Substituting above value in equation (3), we get :

⇒ AD² = AC² + CD.BD

Hence, proved that AD² = AC² + BD.CD

In triangle ABC, angle A = 90°, CA = AB and D is a point on AB produced. Prove that :

DC² - BD² = 2AB.AD.

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle ACD,

By pythagoras theorem,

⇒ CD² = AC² + AD²

⇒ CD² = AC² + (AB + BD)²

⇒ CD² = AC² + AB² + BD² + 2.AB.BD .........(1)

In right angle triangle ABC,

By pythagoras theorem,

⇒ BC² = AC² + AB²

⇒ BC² = AB² + AB² (Since, CA = AB)

⇒ BC² = 2AB²

⇒ AB² = $\dfrac{1}{2}$ BC² .........(2)

Substituting value of AB² from equation (2) in (1), we get :

⇒ CD² = AC² + $\dfrac{1}{2}BC^2$ + BD² + 2.AB.BD

⇒ CD² - BD² = AB² + $\dfrac{1}{2}BC^2$ + 2.AB.BD [As, AC = AB]

⇒ CD² - BD² = AB² + AB² + 2.AB.(AD - AB) [From equation (2)]

⇒ CD² - BD² = 2AB² + 2.AB.AD - 2.AB²

⇒ CD² - BD² = 2.AB.AD

Hence, proved that CD² - BD² = 2.AB.AD

In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that :

BD² - CD² = 2CD × AD

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right-angled triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ AD² = AB² - BD² .......(1)

From figure,

⇒ AC = AD + DC

Squaring both sides, we get :

⇒ AC² = (AD + DC)²

⇒ AC² = AD² + DC² + 2AD.DC

⇒ AC² = AB² - BD² + DC² + 2AD.DC [From equation (1)]

Substituting AB = AC, in above equation :

⇒ AC² = AC² - BD² + DC² + 2AD.DC

⇒ AC² - AC² + BD² - DC² = 2AD.DC

⇒ BD² - DC² = 2CD × AD.

Hence, proved that BD² - DC² = 2CD × AD.

In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1 : 3.

Prove that : 2AC² = 2AB² + BC².

Answer

Given,

D divides BC in the ratio 1 : 3.

⇒ BD : DC = 1 : 3

Let BD = x and DC = 3x

From figure,

⇒ BC = BD + DC = x + 3x = 4x.

$\Rightarrow \dfrac{BD}{BC} = \dfrac{x}{4x} \\[1em] \Rightarrow \dfrac{BD}{BC} = \dfrac{1}{4} \\[1em] \Rightarrow BD = \dfrac{1}{4}BC. \\[1em] \Rightarrow \dfrac{CD}{BC} = \dfrac{3x}{4x} \\[1em] \Rightarrow \dfrac{CD}{BC} = \dfrac{3}{4} \\[1em] \Rightarrow CD = \dfrac{3}{4}BC.$

In right angle triangle ADC,

By pythagoras theorem,

⇒ AC² = AD² + CD² ............(1)

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD² ............(2)

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

⇒ AC² - AB² = AD² + CD² - (AD² + BD²)

⇒ AC² - AB² = CD² - BD²

Substituting value of BD and CD in above equation, we get :

$\Rightarrow AC^2 - AB^2 = \Big(\dfrac{3}{4}BC\Big)^2 - \Big(\dfrac{1}{4}BC\Big)^2 \\[1em] \Rightarrow AC^2 - AB^2 = \dfrac{9}{16}BC^2 - \dfrac{1}{16}BC^2 \\[1em] \Rightarrow AC^2 - AB^2 = \dfrac{8}{16}BC^2 \\[1em] \Rightarrow AC^2 - AB^2 = \dfrac{1}{2}BC^2 \\[1em] \Rightarrow 2(AC^2 - AB^2) = BC^2 \\[1em] \Rightarrow 2AC^2 - 2AB^2 = BC^2 \\[1em] \Rightarrow 2AC^2 = 2AB^2 + BC^2.$

Hence, proved that 2AC² = 2AB² + BC².

In the given figure, AB = 16 cm, BC = 12 cm and CA = 6 cm; find the length of CD.

Answer

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

Let CD be x cm.

In right angle triangle ACD,

By pythagoras theorem,

⇒ AC² = AD² + CD²

⇒ 6² = AD² + x²

⇒ 36 = AD² + x²

⇒ AD² = 36 - x² ........(1)

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 16² = AD² + (BC + CD)²

⇒ 256 = 36 - x² + (12 + x)² [From equation (1)]

⇒ 256 = 36 - x² + 12² + x² + 2(12)x

⇒ 256 = 36 + 144 + 24x

⇒ 256 = 180 + 24x

⇒ 24x = 256 - 180

⇒ 24x = 76

⇒ x = $\dfrac{76}{24} = \dfrac{19}{6} = 3\dfrac{1}{6}$ cm.

Hence, CD = $3\dfrac{1}{6}$ cm.

In a quadrilateral ABCD, ∠A + ∠D = 90°, prove that : AC² + BD² = AD² + BC².

Answer

Produce AB and DC such that they meet at point E.

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In △ AED,

⇒ ∠A + ∠D + ∠E = 180°

⇒ 90° + ∠E = 180°

⇒ ∠E = 180° - 90° = 90°.

By pythagoras theorem,

⇒ AD² = AE² + DE² .........(1)

In △ BEC,

By pythagoras theorem,

⇒ BC² = BE² + CE² .........(2)

In △ AEC,

By pythagoras theorem,

⇒ AC² = AE² + CE² .........(3)

In △ BED,

By pythagoras theorem,

⇒ BD² = BE² + DE² .........(4)

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

⇒ AD² + BC² = AE² + DE² + BE² + CE²

⇒ AD² + BC² = (AE² + CE²) + (BE² + DE²)

⇒ AD² + BC² = AC² + BD².

Hence, proved that AC² + BD² = AD² + BC².