Question

Prove that in a cyclic trapezium, angles at the base are congruent.​

Answer 1

let ABCD be the cyclic trapezium with AB IICD.
thru' C draw CE parallel to AD meeting AB in E.
So
AECD is a parallelogram.
so
angle D=angle AEC.... opp angles of a parallelogram are equal....(i)
but
angle D+angle ABC=180... opp angles of a cyclic quadr are supplementary....(ii)
from (i) and (ii)
angle AEC+angle ABC=180
but
angle AEC+angle CEB= 180...linear pair
so
angle ABC= angle CEB ..(iii)
so
CE=CB... sides opp equal angles are equal.(iv)
but
CE=AD...opp sides of parallelogram AECD.
from (iv) we get
AD=CB
Thus cyclic quadri ABCD is isoceles.
this proves the first part of the question.
now,
join AC and BD, the diagonals.
in triangles DAB and CBA,
AD=CB...proved before
AB=AB common
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.
so
AD=CB... CPCT
hence proved.

Answer 2

Answer:

the triangles between same parallel lines area are equal.......