Prove: if the four sides of a quadrilateral are equal, the quadrilateral is a rhombus. given: ab = bc = cd = da prove: abcd is rhombus

Answer with explanation:

Given: A Quadrilateral A B CD , in which , AB=BC=CD=DA.

To Prove: Quadrilateral A B CD is Rhombus.

Proof:

⇒Properties of Rhombus

1.All sides are equal.

2.Diagonals bisect each other at right angles.

3.Opposite Angles equal.

 In Quadrilateral AB CD

Join AC.

In Δ ADC and Δ ABC

AD=BC

AB=CD

Diagonal AC is Common between two triangles.

⇒ Δ ADC ≅ Δ ABC-----[SSS]

1.---∠B=∠D→→→→[C P C T]

2.--∠D AC=∠B CA--------[C PCT]

Similarly, If you will join B and D, and consider  Δ A B D and ΔC B D, then

    Δ A B D ≅ ΔC B D------[S SS]

And ,3.→→ ∠A=∠C------[C P C T]

4.→→∠BAC=∠DCA------[C P C T]

Opposite angles of Quadrilateral are equal.

Now, we will prove that Diagonals of Rhombus bisect each other at right angle.

Let diagonal AC and BD, intersect at O.

In , Δ A OB and ΔD O C

 AB=CD----[given]

⇒∠ A OB = ∠D O C-------[Vertically Opposite Angles]

⇒∠B AC=∠D CA------Proved at 4

⇒⇒ Δ A OB ≅ ΔD O C------[AAS]

5.→A O=O C------[C PCT]

6.→BO=OD--------[C P CT]

In , Δ A OB and ΔB O C

AB=BC------[Given]

A O=O C----Proved at 5

Segment BO is common.

⇒⇒Δ A OB ≅  ΔB O C-----[S S S]

7.→→∠AOB=∠BOC-------[C P CT]

Now, ∠AOB+∠BOC=180°----[Linear Pair]

→2 ∠BOC=180°

Dividing both sides by ,2 we get

→∠BOC=90°

Which shows that, ∠AOB=∠BOC=∠COD=∠DOA=90°

That is Diagonals Bisect each other at right angles.

Hence ,⇒⇒→→ Quadrilateral ABCD is a rhombus, as we have proved all the three Properties.

r = 78539458390583095803458309458039485093485 or just 3

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