Yes, when I first encountered this one, I spent over 3 minutes trying for a formal solution before seeing that it needs to be greater than 35 😦 And the other method I noticed only on revisiting the problem later – at the second try it struck me quite quickly 😛 Very tricky one to visualise.
CAT-holics 
That was a deadly problem… one of the best u have posted…
Yes, when I first encountered this one, I spent over 3 minutes trying for a formal solution before seeing that it needs to be greater than 35 😦 And the other method I noticed only on revisiting the problem later – at the second try it struck me quite quickly 😛 Very tricky one to visualise.
regards
J
Seriously awesome!
(Y)
This is a very good problem . Is there another way to solve this problem ?
Almost certainly, yes. Most problems would have multiple possible ways of solving. Try and find one!
regards
J
what do you think about my solution ?
Here is how I solved this question:
A( ) — Denotes angle
S( ) — Denotes side
Construction : Join BD
Obseration : A( CBD ) = A( CBD )
So,
A( CBD ) = [ 180 – (y+20) ] / 2 = ( 160 – y ) / 2
A( CDB ) = [ 180 – (y+20) ] / 2 = ( 160 – y ) / 2
A( ABD ) = 140 – A( CBD) = (120 + y) / 2
A( ADB ) = 110 – A( CDB ) = ( 60 + y ) / 2
A( BAD ) = x + 20
Assumption : y S(DC) > S(AD)
Also , in triangle ABD we know that A( ABD ) > A( ADB )
(2) Therefore , in triangle ABD ==> S(AD) > S(AB)
But , we know that S(DC) = S(AB) . Hence , (1) and (2) contradicts each other that means our assumption is wrong .
Now checking values of x and y for the following cases:
y x Compare
20 50 y < x
30 40 y x
So, the correct option is (3) .
There was a formatting problem above .
Assumption is y S(DC) >S(AD)
Assumption y is LESS than x . (1) in triangle ADC side DC is GREATER than side AD