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zaphod Group TitleBest ResponseYou've already chosen the best response.1
http://screencast.com/t/CciNQSaVQjmD
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.1
@ParthKohli @AccessDenied @Aadarsh
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
easy 1minute I am answering u now
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Angle CDX = 56, so, 1/2 angle CDX = angle XDB = 28 dehrees
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.1
okay thanks :)
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
they never say CDX = XDB
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
so you have to prove by complimentary angles
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.1
because it is a parellelogram we can prove right?
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
i used equal angles to prove it
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.1
show me ur method?
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
Given, CD is parallel to EB, so angle CDX = angle EDX = 56 degrees. So, angle DXB = 180  56 = 124 degrees
 2 years ago

Aadarsh Group TitleBest ResponseYou've already chosen the best response.1
So, angle XBD = 180  (124+28) = 28 degrees = angle BDX Hence, traingle BDX is isosceles
 2 years ago

ninhi5 Group TitleBest ResponseYou've already chosen the best response.0
EB // DC => BDC = EBD EBD = EAD => EA // BD => AEB = ADB => EBD = EBD => BDX isosceles
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.1
thanks once again guyz :D
 2 years ago
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