Question

If the diagonals of a quadrilateral bisect each other at right angles, show that it is a rhombus.

Answer 1

Sol:

Given that ABCD is a square.

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

Proof:

(i) In a Δ ABC and Δ BAD,

AB = AB ( common line)

BC = AD ( opppsite sides of a square)

∠ABC = ∠BAD ( = 90° )

Δ ABC ≅ Δ BAD ( By SAS property)

AC = BD ( by CPCT).

(ii) In a Δ OAD and Δ OCB,

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:

Hi mate here is the answer:--✍️✍️

Question:-✔️✔️

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:-✔️✔️

To Prove:

If diagonals of a quadrilateral bisect at 90º, it is a rhombus.

Definition of Rhombus:

A parallelogram whose all sides are equal.

Given:

Let ABCD be a quadrilateral whose diagonals bisect at 90º.

In ΔAOD and ΔCOD,

In ΔAOD and ΔCOD,OA = OC (Diagonals bisect each other)

In ΔAOD and ΔCOD,OA = OC (Diagonals bisect each other)∠AOD = ∠COD (Given)

In ΔAOD and ΔCOD,OA = OC (Diagonals bisect each other)∠AOD = ∠COD (Given)OD = OD (Common)

∆AOD congruent ∆ ΔCOD (By SAS congruence rule)

AD = CD ..................(1)

Similarly,

AD = AB and CD = BC ..................(2)

From equations (1) and (2),

AB = BC = CD = AD

Since opposite sides of quadrilateral

ABCD are equal, it can be said that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that

ABCD is a rhombus

Hence, Proved.

Hope it helps you ❣️☑️☑️

Step-by-step explanation: