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.