The two diagonals of an inscribed quadrilateral meet at the center of a circle.
the quadrilateral must be a __
a. square
b. rectangle
c. rhombus
d. trapezoid

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

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step-by-step explanation:

i hope this is what you were looking for.

see attached graph