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.
