Math, asked by YOGESHmalik025, 1 year ago

Give this attachment answer plz

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Answered by nsharma28
2

Step-by-step explanation:

in triangle age

ab+bf = ac+ce ( given all equal)

af = ae

angle AFE = angle aef

angle AFE = 6x°

as bc || Fe with bf as transversal

angle cbf +angle bfe = 180° ( co int angles)

180°-7x-11+6x = 180

-7x+6x = 180-180+11

-x = 11

x = 11°

7x-11 = 7*11-11

= 77-11

= 66°

angle cbf = 180°-66°

= 114°

angle abc = angle acb

angle acb = angle ecd

= 66°

angle ced = 5x-7

= 48°

in triangle ced

ced+ecd+edc = 180°

48°+ 66° + edc = 180°

edc = 180-114

edc = 66°

as edc = ecd

ed=ec

bf= ed ...(1)

angle fbc + angle edc = 180

180-(7*11-11) + 66 = 180

180-(77-11) + 66 = 180

180-66+66 = 180

180= 180

so angle fbc = angle edc

bf || Ed ( converse of co int angles theorem). ... (2)

from (1) and (2)

bf || ed

bf = ed

so bdef is a parallelogram

hence proved !!!

very lengthy

Answered by Anonymous
0

\huge\blue{Answer}

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In triangle AGE

AB+BF = AC+CE ( given all equal)

af = ae

angle AFE = angle AEF

angle AFE = 6x°

as BC || FE with BF as transversal

angle CBF +angle BFE = 180° ( co int angles)

180°-7x-11+6x = 180

-7x+6x = 180-180+11

-x = 11

x = 11°

7x-11 = 7*11-11

= 77-11

= 66°

angle CBF = 180°-66°

= 114°

angle ABC = angle ACB

angle ACB = angle ECD

= 66°

angle CED = 5x-7

= 48°

in triangle CED

CED+ECD+EDC = 180°

48°+ 66° + EDC = 180°

EDC = 180-114

EDC = 66°

as EDC = ECD

ED=EC

BF= ED ...(1)

angle FBC + angle EDC = 180

180-(7*11-11) + 66 = 180

180-(77-11) + 66 = 180

180-66+66 = 180

180= 180

so angle FBC = angle EDC

BF || ED ( converse of co int angles theorem). ... (2)

from (1) and (2)

BF || ED

BF = ED

so BDEF is a parallelogram

<marquee><font color= "red">Hope it helps you mark as brainliest please

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