Question

(111) The diagonals of a quadrilateral intersect each other at right angle. Prove that the
sum of the squares of opposite sides is equal.​

Answer 1

Answer:

Step-by-step explanation:

Given : ABCD is any quadrilateral.

Diagonals AC and BD intersect at right angle at point O.  

To Prove : $AB^{2} + CD^2 = AD^2 + BC^2$

Proof :

In  ΔAOB, ∠AOB = 90°

∴  by Pythagoras theorem,

$AB^2 = AO^2+OB^2$ . . . . .(1)

In ΔCOD, ∠COD = 90°

∴  by Pythagoras theorem,

$CD^2 = CO^2+DO^2$   . . . . .(2)

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

$AB^2+CD^2 = AO^2+OB^2+CO^2+OD^2$  . . . . .(3)

In ΔAOD, ∠AOD = 90°

∴  by Pythagoras theorem,

$AD^2= AO^2+OD^2$ . . . .(4)

ΔCOB,∠COB = 90°

∴  by Pythagoras theorem,

$BC^2=BO^2+OC^2$  . . . . .(5)

Adding (4) and (5), we get,

$AD^2+BC^2=AO^2+OD^2+BO^2+OC^2$  . . . .(6)

∴ from (3) and (6), we get,

$AB^2+CD^2=AD^2+BC^2$

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Answer 2

Answer:

Step-by-step explanation:

Given : ABCD is any quadrilateral.

Diagonals AC and BD intersect at right angle at point O.

To Prove : AB^{2} + CD^2 = AD^2 + BC^2AB

2

+CD

2

=AD

2

+BC

2

Proof :

In ΔAOB, ∠AOB = 90°

∴ by Pythagoras theorem,

AB^2 = AO^2+OB^2AB

2

=AO

2

+OB

2

. . . . .(1)

In ΔCOD, ∠COD = 90°

∴ by Pythagoras theorem,

CD^2 = CO^2+DO^2CD

2

=CO

2

+DO

2

. . . . .(2)

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

AB^2+CD^2 = AO^2+OB^2+CO^2+OD^2AB

2

+CD

2

=AO

2

+OB

2

+CO

2

+OD

2

. . . . .(3)

In ΔAOD, ∠AOD = 90°

∴ by Pythagoras theorem,

AD^2= AO^2+OD^2AD

2

=AO

2

+OD

2

. . . .(4)

ΔCOB,∠COB = 90°

∴ by Pythagoras theorem,

BC^2=BO^2+OC^2BC

2

=BO

2

+OC

2

. . . . .(5)

Adding (4) and (5), we get,

AD^2+BC^2=AO^2+OD^2+BO^2+OC^2AD

2

+BC

2

=AO

2

+OD

2

+BO

2

+OC

2

. . . .(6)

∴ from (3) and (6), we get,

AB^2+CD^2=AD^2+BC^2AB

2

+CD

2

=AD

2

+BC

2

Hope This Helps You