Question

show that the diagonals of square are equal and bisects each other at right angle​

Answer 1

Answer:

Step-by-step explanation:

Given that ABCD is a square.

To prove : AC=BD and AC and BD bisect each other at right angles.

Proof:  

(i) In a ΔABC and ΔBAD,

AB=AB ( common line)

BC=AD ( opppsite sides of a square)

∠ABC=∠BAD ( = 90° )

ΔABC≅ΔBAD( By SAS property)

AC=BD ( by CPCT).

(ii) In a ΔOAD and ΔOCB,

AD=CB ( opposite sides of a square)

∠OAD=∠OCB ( transversal AC )

∠ODA=∠OBC ( transversal BD )

ΔOAD≅ΔOCB (ASA property)

OA=OC ---------(i)

Similarly OB=OD ----------(ii)

From (i) and (ii)  AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB=OD ( from (ii) )

BA=DA

OA=OA  ( common line )

ΔAOB=ΔAOD----(iii) ( by CPCT

∠AOB+∠AOD=180°  (linear pair)

2∠AOB=180°

∠AOB=∠AOD=90°

∴AC and BD bisect each other at right angles.

Answer 2

Answer:

Hope it helps :)

Step-by-step explanation:

Given- ABCD is a square, diagonals AC and BD intersect at O.

To Prove-1. AC=BD.

2. AO=OC.

3. ANGLE AOB=90 degree.

Proof- In triangle ABC and triangle BAD

AB=AB(common)

Angle ABC=Angle BAD(each 90 degree)

BC=AD(given)

By SAS Criterion, triangle ABC is congruent to triangle BAD.

AC=BD(by CPCT)

Hence, diagonals are equal.

In triangle AOB and triangle COD

Angle BAO=Angle DCO(Alternate Interior Angles)

Angle AOB=Angle COD(Vertically Opposite Angles)

AB=CD(given)

By AAS criterion, triangle AOB is congruent to triangle COD

AO=CO(by CPCT)

Hence, diagonals bisect each other.

In triangle AOB and triangle COB

OB=OB(given)

AO=CO(diagonals are bisected)

AB=CB(sides of the square)

By SSS criterion, triangle AOB is congruent to triangle COB

angle AOB=angle COB

angle AOB+angle COB=180 degree(Linear Pair)

angle AOB=90 degree

Hence, diagonals bisect each other at right angles.