Question

If non-parallel sides of trapezium are equal, Prove that the trapezium is cyclic.

Answer 1

Let ABCD be a trapezium where AB//CD and AD=BC

Construction: Draw Perpendicular to AB and CM perpendicular to AB

From triangles ALD and BMC we have

AD=BC(given)

DL=CM(distance between parallel sides)

angle ALD = angle BMC (90 degree)

Therefore by RHS congruence criterion, triangle ALD is congruent to triangle BMC

hence angle DAL = angle CBM (C.PC.T) 1)

Since AB// CD

angle DAL + angle ADC = 180 degree (sum of adjacent interior angles is supplementary)

implies angle CBM + angle ADC = 180 degree(from 1)

Implies ABCD is a cyclic trapezium(SUM of opposite angles is supplementary)

Answer 2

consider a trapezium ABCD with AB parallel CD  and BC =AD 
draw AM perpendicular CD and BM perpendicular  to CD 
in triangle AMD  and BMC  
AD = BC  given 
angle AMD =angle BMC  by construction each angle at 90 degree 
AM = BM  perpendicular distance between two lines is same 
triangle AMD  congurant to triangle BNC by  (RHS rule )
angle ADC = angle BCD by (cpct) ----------1 
angle BAD and angle  ADC  are on same side of transvercal AD
angle  ADC + angle BAD = 180 degree ---------2
angle BAD + angle BCD = 180 degree  ( using 1st equation )
this equation shows the opposite angles are supplementry 
 THERE FORE  ABCD IS A CYCLIC QUADRILATERALS 

HENCE PROVED