IN THE GIVEN FIGURE <ABC=30°,<EDF=(40-X)° AND <ADE = (13X+20) . SHOW THAT BC PARALLEL TO DE
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hello here is your answer by Sujeet yaduvanshi ☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝
Given : <ABC= 30, <EDF= (40-x) degree and <ADE= (13x+20)
To prove : BC is parallel to DE
Proof:
<ABC+ < DBC = 180 (Linear pair axiom)
30 + <DBC= 180
< DBC = 180-30
= 150 degrees
also, < ADE + < EDF = 180 (Linear pair axiom)
(13x+20)+(40-x) = 180
12x +60 = 180
12x = 180 - 60
12x = 120
x = 120/12 = 10
x = 10
Now substitute the value of x in both the terms ...
<ADE = 13x+20 : where x = 10 <EDF = 40-x
13*10+20 = 150 degrees & 40-10 = 30 degrees...
As seen above ,
< DBC = < ADE (Alternate angles are equal)
Hence, we can say that - BC is parallel to DE
that's all
Given : <ABC= 30, <EDF= (40-x) degree and <ADE= (13x+20)
To prove : BC is parallel to DE
Proof:
<ABC+ < DBC = 180 (Linear pair axiom)
30 + <DBC= 180
< DBC = 180-30
= 150 degrees
also, < ADE + < EDF = 180 (Linear pair axiom)
(13x+20)+(40-x) = 180
12x +60 = 180
12x = 180 - 60
12x = 120
x = 120/12 = 10
x = 10
Now substitute the value of x in both the terms ...
<ADE = 13x+20 : where x = 10 <EDF = 40-x
13*10+20 = 150 degrees & 40-10 = 30 degrees...
As seen above ,
< DBC = < ADE (Alternate angles are equal)
Hence, we can say that - BC is parallel to DE
that's all
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