Question

prove that an isosceles trapezium is always cyclic.

Answer 1

An isosceles trapezium is always cyclic.

Step-by-step explanation:

To show that any given quadrilateral is cyclic, we must show that its opposite angles are supplementary (that is, they add up to 180°).

Consider an isosceles trapezium PQRS, where  PQ and RS are parallel and PS = QR.

We need to prove that ∠QPS + ∠QRS = 180° and ∠PSR + ∠PSR = 180˚.

Proof: Construct two perpendiculars, PF and QE, from segment PQ to segment RS.

Now, in ΔPSF and ΔQRE,

• ∠PFS = ∠QER (by construction - both are right angles at the feet of the perpendiculars)
• PS = QR (property of trapezium)
• PF = QE ( perpendicular distance between two parallel lines remains same irrespective of the point of measurement along the line )

Thus, ΔPSF ≅ ΔQRE ( Right angle - Hypotenuse - Side congruency.)

 Now,

∠PSF = ∠QRE ( By CPCT i.e. Corresponding Parts of Congruent Triangles )

Since ∠PSF = ∠PSR and ∠QRE = ∠QRS,

∠PSR = ∠QRS (1)

Also,

∠RQE = ∠SPF ( By CPCT )

Adding the right angles ∠PQE and ∠QPF to the above angles,

∠RQE + ∠QPF = ∠RQE + ∠PQE

Thus, ∠PQR = ∠QPS (2)

So, adding 1 and 2,

∠PSR + ∠PQR = ∠QRS + ∠QPS

We know that the sum of all the angles in a quadrilateral is 360˚,

∴ ∠PSR + ∠PQR + ∠QRS + ∠QPS = 360˚

⇒ 2 (∠PSR + ∠PQR) = 2 (∠QRS + ∠QPS) = 360˚

⇒ ∠PSR + ∠PQR =∠QRS + ∠QPS = 180˚

Since the opposite angles are supplementary, an isosceles trapezium is a cyclic quadrilateral.

Hence proved.