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Isosceles Trapeziums are Con-Cyclic

**Problem** : We need to prove that any isosceles trapezium is con-cyclic.

**Solution** :Here’s an isosceles trapezium with sides PQ and RS are parallel and PS = QR. A quadrilateral is cyclic if it’s opposite angles are supplementary (i.e. they add up to 180˚). So , we need to prove that QPS + QRS = 180˚ and PSR + PQR = 180˚.

Let us construct two perpendiculars, PU and QT, from point P and Q to segment RS.
Now, in ΔPSU and ΔQRT

PUS = QTR by construction - both are right angles

PS = QR Given trapezium is isosceles

PU = QT perpendicular distance between two parallel lines

Thus, ΔPSU and ΔQRT are congruent by Right angle-Hypotenuse-Side (RHS) congruency rule

Now angles ,
PSU = QRT Corresponding Parts of Congruent Triangles (CPCT)

Therefore ,angle PSR and angle QRS are equal. -------- (1)

Also, angle RQT is equal to angle SPU by CPCT. Adding right angles UPQ and TQP to the above angles, we get

RQT + TQP = SPU + UPQ

Thus, angle RQP is equal to SPQ -------- (2)

Adding equations 1 and 2 we get the following relations for angles

PSR + RQP = QRS + SPQ --------(3)

Since the sum of all the angles in a quadrilateral is 360˚,from equation (3)

PSR + RQP + QRS + SPQ = 360˚

2 (PSR + RQP) = 2 (QRS + QPS) = 360˚

PSR + RQP =QRS + QPS = 180˚

**Since the opposite angles are supplementary, it can be concluded that an isosceles trapezium is a cyclic quadrilateral.**
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