Question

prove that diagonals of the square is equal and perpendicular to each other​

Answer 1

Answer:

(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

Step-by-step explanation:

Answer 2

Answer:

The diagonals of a square are perpendicular bisectors of one another. As a result, Their intersection forms four right angles, and each diagonal is split into two congruent pieces. Therefore, if given the length of a diagonal, the length of one segment of that diagonal is half of the length of the entire diagonal.

Step-by-step explanation:

let

ABCD is a square with diagonals AC and BD intersecting each other at point O.

To prove: AC = BD and ZAOB =

90°

Proof:

In ADAB and ACBA,

AD = BC

...(Sides of a

square)

ZDAB = ZCBA

... (Each 90) ...(Each

AB = AB

...(Common)

⇒ ADAB ACBA... (SAS

Congruence)

⇒AC = BD

...(C.P.C.T)

In AAOB and ABOC,

AD = BC

...(Sides of a

square)

OB = OB

...(Common)

OA = OC

...(Diagonals of a

square bisect each other)

⇒AAOBABOC ...(SSS

Congruence)

⇒ ZAOB = ZBOC ...(C.P.C.T)

But, ZAOB + ZBOC = 180°

(Linear pair)

180° 2 ⇒ ZAOB = ZBOC = 90°

.. AC = BD ⇒ Diagonals are

equal

And, ZAOB = 90° ⇒ Diagonals

are perpendicular to each other.