Question

Show that the diagonals of square are equal and perpendicular to each other

Answer 1

(draw a square ABCD with diaagonals AC and BD)
Given:ABCD is a square
To prove:AC=BD and AC and BD bisect each other at right. angles

Proof:In triangle ACB and triangle BAD

AB=AB (common side)
angle ABC=angle BAD(=90degree)
BC=AD(opposite sides of a square)

Triangle ACB congruent BAD(SAS criteria)
AC=BD by(CPCT)

in triangle OAD and OBC

angle OAB=OCB(AC transversal)
AD=CB(opposite sides of square)
angle ODA=OBC(BD transversal)

OA=OC........

Answer 2

Given :- ABCD is a square.

To proof :- AC = BD and AC ⊥ BD

Proof :- In △ ADB and △ BCA

AD = BC [ Sides of a square are equal ]

∠BAD = ∠ABC [ 90° each ]

AB = BA [ Common side ]

△ADB ≅ △BCA [ SAS congruency rule ]

⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]

In △AOB and △AOD

OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]

AB = AD [ Sides of a square are equal ]

AO = AO [ Common side ]

△AOB ≅ △ AOD [ SSS congruency rule ]

⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]

∠AOB + ∠AOD = 180° [ Linear pair ]

∠ AOB = ∠AOD = 90°

⇒ AO ⊥ BD

⇒ AC ⊥ BD

Hence proved, AC = BD and AC ⊥BD