Answer with explanation:
Draw a Quadrilateral , A B CD, inside a circle having Center , O.The Diagonals AC and B D, intersect each other at O.
Let Radius of Circle = r unit
→In Δ A OB and Δ COD
AO=OC=r
BO=OD=r
∠AOB=∠COD⇒Vertically Opposite Angles
→Δ A OB ≅ Δ COD ⇒[SAS]
By, C PCT
AB=CD
x=u
Similarly,by ,S AS,⇒ ΔAOD ≅ ΔBOC
So, By C PCT,
AD=BC
y=v
In a cyclic Quadrilateral, sum of Opposite angles are Supplementary.
⇒∠A+∠C=180°
→x+y+u+v=180°
→2 x+2 y=180°
→ x+y=90°
∠A=90°
⇒∠A=∠B=∠C=∠D=90°
So, in Quadrilateral , A B CD
⇒∠A=∠B=∠C=∠D=90°
as well as, AB=CD and AD=BC.
Opposite sides are equal and each interior angle has measure 90°.
The quadrilateral must be a
B. Rectangle