Question

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see adjacent figure).
Show that angle A> angle C and angle B> angle D

Answer 1

Answer:

AB- smallest side

CD-Longest side

in triangle A

Answer 2

B is the opposite angle of the larger side CD

D is the opposite angle of comparatively smaller side BC

so angle B > D

consider ABCD is a cyclic quadrilateral

angle formed in the smaller arc BAD will be larger than the angle formed in the larger arc BCD

so A >C

I know it's just a simple way

not well explained but I think u will get the point...