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zaphodBest ResponseYou've already chosen the best response.1
http://screencast.com/t/CciNQSaVQjmD
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
@ParthKohli @AccessDenied @Aadarsh
 one year ago

AadarshBest ResponseYou've already chosen the best response.1
easy 1minute I am answering u now
 one year ago

AadarshBest ResponseYou've already chosen the best response.1
Angle CDX = 56, so, 1/2 angle CDX = angle XDB = 28 dehrees
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
they never say CDX = XDB
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
so you have to prove by complimentary angles
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
because it is a parellelogram we can prove right?
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
i used equal angles to prove it
 one year ago

AadarshBest 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
 one year ago

AadarshBest ResponseYou've already chosen the best response.1
So, angle XBD = 180  (124+28) = 28 degrees = angle BDX Hence, traingle BDX is isosceles
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
EB // DC => BDC = EBD EBD = EAD => EA // BD => AEB = ADB => EBD = EBD => BDX isosceles
 one year ago

zaphodBest ResponseYou've already chosen the best response.1
thanks once again guyz :D
 one year ago
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