Give this attachment answer plz
Answers
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
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