Question

(i) चक्रीय समलंब चौकोन हा समद्विभुज समलंब चौकोन असतो, हे सिद्ध
करा.​

Answer 1

Given : चक्रीय समलंब चौकोन हा समद्विभुज समलंब चौकोन असतो,

cyclic trapezium is an isosceles trapezium

To Fine : सिद्ध करा.​

prove

Solution:

Let say ABCD is a cyclic trapezium  where

AD || BC

Hence we need to show that AB = CD for it to be an isosceles trapezium

as its  cyclic => sum of opposite angles = 180°

Lets extend AD  and draw a line parallel to AB passing through C and intersecting  extended AD at  E.

AB || CD  and  CE || AB

hence ABCE is a parallelogram

=> AB = CE  opposite sides are equal

in parallelogram opposite angles are equal

Hence ∠B = ∠E

∠B + ∠D  = 180°    ( cyclic Quadrilateral )

∠D + ∠CDE  = 180°   linear Pair

=> ∠E = ∠CDE

in ΔCDE

∠E = ∠CDE

=> CD = CE

  AB = CE    ( opposite sides of parallelogram )

=> AB = CD

Hence trapezium is isosceles trapezium

QED

Hence proved

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Answer 2

Step-by-step explanation:

(i) चक्रीय समलंब चौकोन हा समद्विभुज समलंब चौकोन असतो, हे सिद्ध

करा.