Respuesta :

Answer:

Proof is given below.

Step-by-step explanation:

Given:

ABCD is a Square.

i.e AB = BC = CD = DA

To Show:

Diagonals of a square are equal and bisect each other.

1.  AC = BD

2. AO = CO and   BO = DO

Proof:

ABCD is a Square  ............{Given}

∴ AB = BC = CD = DA ...........{All sides are equal of a Square ( 1 )}

∴ m∠ A = m∠ B = m∠ C = m∠ D = 90° ..{Measure of each vertex angle is 90° ( 2 )}

Now ,In  Δ ADC and Δ BCD

AD ≅ BC     ……….{From ( 1 )}

∠ ADC ≅ ∠ BCD = 90°     …………..{ From ( 2 )}

DC ≅ DC      ……….{Reflexive Property}

Δ ADC ≅ Δ BCD ….{Side-Angle-Side test}

∴ ∠ DAC≅ ∠ CBD...{corresponding angles of congruent triangles (c.p.c.t)( 3 )}

∴ AC ≅ BD  ......{corresponding sides of congruent triangles (c.p.c.t) }

∴ AC ≅ BD  ..... Proved ( 1 ) Diagonals are Equal

Now,In  Δ AOD and Δ BOC

∠ DAO ≅ ∠ CBO     …………..{From ( 3 ) c.p.c.t}

∠ DOA ≅ ∠ COB       ……….{Vertically Opposite Angles}

AD ≅ BC     ……….{From ( 1)}

Δ AOD ≅ Δ BOC ….{ by Angle-Angle-Side test}

∴ AO ≅ OC  ......{corresponding sides of congruent triangles (c.p.c.t) }

∴ DO ≅ BO  ......{corresponding sides of congruent triangles (c.p.c.t) }

i .e Diagonal bisect each other....... Proved..

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Answer:

Solution

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.

solution

Step-by-step explanation:

Solution

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.

solution

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