prove that a cyclic trapezium is isosceles and its diagonals are equal....
Asked by harshitalohia | 7th Feb, 2010, 10:50: AM
let ABCD be the cyclic trapezium with AB IICD.
thru' C draw CE parallel to AD meeting AB in E.
AECD is a parallelogram.
angle D=angle AEC.... opp angles of a parallelogram are equal....(i)
angle D+angle ABC=180... opp angles of a cyclic quadr are supplementary....(ii)
from (i) and (ii)
angle AEC+angle ABC=180
angle AEC+angle CEB= 180...linear pair
angle ABC= angle CEB ..(iii)
CE=CB... sides opp equal angles are equal.(iv)
CE=AD...opp sides of parallelogram AECD.
from (iv) we get
Thus cyclic quadri ABCD is isoceles.
this proves the first part of the question.
join AC and BD, the diagonals.
in triangles DAB and CBA,
angle ADB= angle ACB.. angles in the same segment of a circle are equal.here AB is the chord.
so triangles DAB and CBA are congruent....SAS rule.
Answered by | 9th Feb, 2010, 04:55: PM
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