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Re: In the figure shown, line segment AD is parallel to line segment BC. [#permalink]
21 Aug 2011, 10:00

LM wrote:

x = 180-y.

Corresponding angle property of parallel lines. Thus A alone is sufficient.

If y=50, then the upper right angle of the left triangle is also 50. How can you calculate x = 180 - y? For this to work, BCD needs to be 180° (so y would be the supplementary of x)

Re: In the figure shown, line segment AD is parallel to line segment BC. [#permalink]
30 Sep 2011, 14:01

Given that BC is parallel to AD, extend both lines to visually see that line AC is a transversal for the parallel lines BC and AD.

a) Thus, as the # of degrees for a straight line is equal to 180, and given that y = 50, by the alternate interior angle rule, the angle adjacent to angle x will also be 50 degrees. And since we've extended line BC beyond point C, angle x will equal 180 (degress in a straight line) minus 50.

b) Same thought process as above. Now, line CD is the transversal between parallel lines BC and AD, again allowing us to find the measure of angle x.

Re: In the figure shown, line segment AD is parallel to line segment BC. [#permalink]
30 Sep 2011, 16:17

baker2145 wrote:

Given that BC is parallel to AD, extend both lines to visually see that line AC is a transversal for the parallel lines BC and AD.

a) Thus, as the # of degrees for a straight line is equal to 180, and given that y = 50, by the alternate interior angle rule, the angle adjacent to angle x will also be 50 degrees. And since we've extended line BC beyond point C, angle x will equal 180 (degress in a straight line) minus 50.

b) Same thought process as above. Now, line CD is the transversal between parallel lines BC and AD, again allowing us to find the measure of angle x.

My answer: D

Although I could be totally off.

What's official answer please?

looks like question is wrong. how did u come up with D?

gmatclubot

Re: In the figure shown, line segment AD is parallel to line segment BC.
[#permalink]
30 Sep 2011, 16:17

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