Question

if non - parellel sides of trapezium are equal, prove that it is cyclic.​

Answer 1

To Prove:

If the non-parallel sides of a trapezium are equal, we need to prove that it is a cyclic quadrilateral.

Concepts used:

1. A trapezium is a quadrilateral with one pair of sides as parallel.
2. A trapezium with the non-parallel sides equal is called Isosceles trapezium.
3. In every isosceles trapezium, the base angles are equal.
4. In every trapezium (isosceles or non-isosceles), adjacent angles are supplementary or their sum is equal to 180 degrees.
5. A cyclic quadrilateral can be inscribed within a circle and its opposite angles are supplementary or have a sum equal to 180 degrees.

Accordingly, let's proceed,

• Let the trapezium be ABCD.

• Here, AB and CD are the parallel sides.

• AC and BD are non-parallel sides.

• As ABCD is an isosceles trapezium (Non parallel sides are equal).

• So, ∠ACD = ∠BDC = x (let both be equal to x degrees)

• ∵ AB is parallel to CD (From figure - ABCD is a trapezium )

• Hence, AC and BD can be considered to be transversals cutting the two parallel lines AB and CD.

• Thus, ∠ ACD + ∠ CAB = 180 °  &   ∠ABD + ∠ BDC = 180 ° (From 4)

• x + ∠ CAB = 180 ° & ∠ABD + x = 180 °

• ∴  ∠CAB = 180 - x   &   ∠ABD = 180 - x

»»› We know that if the sum opposite angles of this trapezium is 180 degrees, then it is cyclic.

»»› So, lets find the sum.

• ∠ ACD + ∠ ABD = ?
• ∠ CAB + ∠ CDB = ?

=> Applying the previously designated values.

• ∠ ACD + ∠ ABD = x + 180 - x

=> ∠ ACD + ∠ ABD =  180 degrees

• ∠ CAB + ∠ CDB = 180 - x + x

=> ∠CAB + ∠ CDB = 180 degrees

=>  ∠ACD and ∠ABD are supplementary

=> ∠CAB and ∠CDB are supplementary

»»› Thus, the opposite angles of this isosceles trapezium are supplementary.

»»› So, it is a cyclic quadrilateral (from 5)

Answer 2

Answer:

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