Question

Prove that isoscles rhombus is a cyclic quadrilateral

Answer 1

Answer:

let the trapezium be ABCD with AB ll CD

then angleA = angleB

angleC = angleD

By ASP of quadrilateral

angleA + angleB + angleC + angleD = 360°

from above

angleA + angleA + angleC + angleC = 360°

2(angleA + angleC)=360°

angleA + angleC = 180°

similarly,

angleB + angleD = 180°

since sum of opposite angle is 180° therefore it is a cyclic Quadrilateral

Answer 2

Given- ABCD is a trapezium with ABIICD and BC=AD(as it is isosceles)
To prove- ABCD is a cyclic quadrilateral 
Construction- Drop perpendiculars AM and BN on DC
Proof-In ΔAMD and ΔBNC
AD=BC(given)
angleAMD=angleBNC=90degree
AM=BN (perpedicular distance between two parallel lines is same)
Therefore, ΔAMD CONGRUENT TO ΔBNC (By RHS congruence rule)
angleADC=angleBCD (CPCT) .... (1)
angleBAD+angleADC=180degree  (angle on the same side of transversal AD) ....
(2)
FROM (1) and (2)
angleBAD+angleBCD=180degree
⇒Opposite angles are supplementary
Therefore, ABCD is a cyclic quadrilateral.
Hence, proved