Question

12. Prove that. (i) diagonals of a rectangle are equal. (ii) diagonals of a square are equal.​

Answer 1

Answer:

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Step-by-step explanation:

i) Given ABCD is a rectangle,

then AC and BD are diagonals,

then in triangle ABC and BCD,

angle b is common angle

BC is common side,

AB = CD

so by SAS congruency,

triangle ABC is congruent to BCD.

so by cpct,

AC = BD

so diagonals are equal

hence proved

ii) 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

Step-by-step explanation:

that is only first I Let ABCD be a rectangle.

Then AC and BD are the diagonals.

To Prove: AC=BD

Proof : In △ABC and △ABD

∠ABC = ∠BAD [90∘]

BC=AD [Opp. Sides]

AB=AB [Common side]

△ABC ≅ △ABD[SAS]

AC=BD

Hence proved.