Question

show that diagonals of a square are equal and bisect each other at right angle​

Answer 1

Answer:

given: ABCD is a square

Diagonals intersect at o

to prove: AC=BD

AO=OC and BO=OD

and angle AOB =BOC=COD=DOA

proof: in ΔABC and ΔDCB

AB=DC (sides of square)

angle ACB=DCB dboth 90°)

BC=BC (common)

ΔABC=ΔDCB by SAS rule

AC=BD by CPCT

Answer 2

Answer:

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.

MARK BRAINLIEST