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given thatabcd ia paralellogram

in triangle afd and triangle bec

af=be(opp sides of a rectangle

ad=bc(opp sides of papallelogram)

df=ce(ab=dc=de+ec

ab=ef=de+df)

afdcongruent to bec(sss congruence)

2

abcd is a rhombus

let acand bd meet at o

oab=ocd

ab=cd

oba=odc

aod congruent to cod

ao=oc(cpct)

aod congruent to cod

similarly we can prove

aodcongruent to cod

from 123 we have

diognals of rhombus divide in to 4 congruent triangles

3

in quadrilateral abcd a+B+C+D=360

1|2c+d+=180-1\2(a+b)

1|2c+1|2d+cod=180

1|2c+1|2d=180-cod

1|2(c+d)=180-cod

from1 2

180-cod=180-1|2(a+b)cod=1|2(a+b)