Question

non-parallel sides of a trapezium are equal, prove that It is Cyclic.

Diagram is also needed.​

Answer 1

★Given:

ABCD is a trapezium where non-parallel sides AD and BC are equal.

★Construction:

DM and CN are perpendicular drawn on AB from D and C respectively.

★To prove:

ABCD is cyclic trapezium.

★Proof:

In △DAM and △CBN,

AD=BC ...Given

∠AMD=∠BNC ...Right angles

DM=CN ...Distance between the parallel lines

△DAM≅△CBN by RHS congruence condition.

Now,

∠A=∠B ...by CPCT

Also, ∠B+∠C=180° ....Sum of the co-interior angles

⇒∠A+∠C=180°

Thus, ABCD is a cyclic quadrilateral as sum of the pair of opposite angles is 180°.

Answer 2

Answer :

Given :

It is given that a trapezium PQRS with parallel sides PQ and SR . Also with non parallel sides QR = PS .

To prove :

Prove that PQRS is a cyclic trapezium .

Construction :

Draw SM perpendicular to PQ and RN perpendicular to PQ .

Proof :

In triangle SMP and triangle RNQ

Angle SMP = Angle RNQ ( each 90° )

QR = PS ( given )

SM = RN ( distance between parallel lines )

Therefore , ∆ SMP $\cong$ ∆ RNQ by RHS rule

$\implies$ Angle SMP = Angle RQN by cpct - i

$\implies$ Angle PSM = Angle QRN by cpct - ii

= 90 - angle PSM = 90 - angle QRN

= angle MSR - angle PSM = angle SRN - angle QRN

= angle PSR = angle SRQ

= Angle S = Angle R - iii

= 180 - angle SPM = 180 - angle RQN

= angle SPQ = angle RQP

= Angle P = Angle Q - iv

In trapezium PQRS ,

= Angle P + Angle Q + Angle R + Angle S = 360

From iv and iii

= Angle Q + Angle Q + Angle S + Angle S = 360

= 2( Angle Q + Angle S ) = 360

= Angle Q + Angle S = $\frac{360}{2}$

= Angle Q + Angle S = 180

Since , sum of opposite angles of trapezium are 180 . Hence proved that PQRS is a cyclic trapezium .