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.

Medians and Isosceles Triangle

If two medians in a triangle are equal in length,then the triangle is isosceles.

Let medians AM = BN in ∆ ABC. Extend each median to AP and BQ so that M and N are the midpoints of AP and BQ, respectively. Hence , AM = MP and BN = NQ. By the property of bisecting diagonals, ABPC and ABCQ are parallelograms. Hence CP and CQ are each parallel and equal to AB. We conclude that C lies on QP and C is the midpoint of QP. 

Now , AM = BN , so  2AM = 2BN  , hence AP = BQ. This shows that ABPQ is a trapezium with equal diagonals. 

It is easy to see that such a trapezium is isosceles. One way to see this is to draw a line through A parallel to diagonal BQ, until it intersects line QP in point L. Thus, ABQL is a parallelogram, so ∠ALQ = ∠ABQ. On the other hand, ∆APL is isosceles since AL = BQ = AP; hence, ∠ ALQ = ∠ APQ. 

Finally, AB || QP implies ∠  APQ = ∠ BAP. We conclude that ∠ BAP = ∠ ABQ, and ∆ ABQ and ∆ BAP are congruent by two equal sides and angles between these sides (SAS Criteria) . Therefore, BP = AQ and our trapezium is isosceles. 

 Hence ∠AQC=∠ BPC. Finally , ∆ACQ and ∆ BCP are congruent by AQ = BP, CQ = CP and ∠ AQC = ∠ BPC (SAS Criteria).  We conclude that AC = BC and our original ABC is isosceles.

Pythagoras Theorem # 4

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