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

2
by pravin3103

2014-12-28T10:54:04+05:30

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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

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)

2014-12-28T10:58:16+05:30

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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