from chapter circles

Asked by ridhisood | 22nd Feb, 2010, 06:58: PM

Expert Answer:

let ABCD be the trapezium such that AB II CD

we have to prove that

if AD=BC, then ABCD is cyclic.

construction:draw a line thru' C parallel to AD meeting AB in E.

then AECD is a parallelogram... as both pairs of opp sides are parallel.

angle ADC=angle AEC.....(i).. opp angles of a parallelogram

now, AD=BC.... given

AD=CE... opp sides of II gram

so,

CE=BC

So,

 

angle CEB=angle CBE

but

 angle CEA+CEB=180... linear pair

so

angle CBE+CEA=180

so from (i)  we get,

angle CDA+CBE=180

 i.e.

 angle CDA+CBA=180

 so quad ABCD is cyclic as one pair of opp angles is supplementary.

 

Answered by  | 25th Feb, 2010, 04:43: PM

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