Question

show that the diagonal of a square are equal and bisect each other at right angle

Answer 1

Name the square as ABCD the diagonals AC and BD meet at O

in triangle AOB and triangle COD
<AOB = <COD (VERTICALLY OPPOSITE ANGLES)
AB = CD (opposite sides of square )
<BAO=<DCO ( alternate interior angles )

So, tr AOB IS CONGRUENT TO tr COD
Hence AO= OC and OB = OD ( corresponding parts of congruent triangles )

Now in tr AOB and tr AOD
AO = AO ( COMMON SIDE)
AB = AD (SIDES OF SQUARE)
OB=OD (PROVED ABOVE)

Therefore tr AOB is congruent to tr AOD (BY SSS criterion of congruency )

So, <AOB = <AOD
Now, <AOB+<AOD=180 (linear pair)
=>2<AOB=180
=> <AOB=90 DEGREE

HENCE PROVED THAT DIAGONALS IN A SQUARE BISECT EACH OTHER PERPENDICULARLY.

Please don't forget to mark it brainliest becoz it took so much time to write it☺☺☺☺☺

Answer 2

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.