Question

Diagonal of cyclic quadrilateral abcd are congruent show that ad=bc and seg ab parallel to seg cd

Answer 1

Seg AB is parallel to Seg CD.

and, AD = BC

Step-by-step explanation:

Given,

ABCD is a quadrilateral in which both the diagonals are congruent,where both the diagonal AC and BD intersect each other at O.

Length of OA = Length of OC

and, Length of OB = Length of OD

Since, Both the diagonals are equal in length.

So,

Length of OA = Length of OC = Length of OB = Length of OD

Hence, ΔOAB, ΔOBC,ΔOCD,ΔODA  are isosceles triangles.

Also, ∠AOB = ∠BOC = ∠COD = ∠DOA = x

As we know that a complete angle is 360°

So, ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

     x+x+x+x=360°

      4x = 360°

       x= $\frac{360}{4}$

         = 90°

⇒∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Also,it is concluded that the all sides of quadrilateral are equal.

So, the given quadrilateral is a square.

⇒ AD = BC = AB =CD

∴, Seg AB is parallel to seg CD.

Answer 2

Given,

ABCD is a quadrilateral in which both the diagonals are congruent,where both the diagonal AC and BD intersect each other at O.

Length of OA = Length of OC

and, Length of OB = Length of OD

Since, Both the diagonals are equal in length.

So,

Length of OA = Length of OC = Length of OB = Length of OD

Hence, ΔOAB, ΔOBC, ΔOCD ,ΔODA  are isosceles triangles.

Also, ∠AOB = ∠BOC = ∠COD = ∠DOA = x.

we know that a complete angle is 360°

So, ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°

     x+x+x+x=360°

      4x = 360°

       x= 360 /4

         = 90°

⇒∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Also,it is concluded that the all sides of quadrilateral are equal.

So, the given quadrilateral is a square.

⇒ AD = BC = AB =CD

∴, Seg AB is parallel to seg CD.

Therefore, Seg AD = Seg BC

( sides of a square.)

Mark me the brainliest.

The figure is given in the picture.