RIGHT OFF THE BAT, MARK SIDES BD AND BC AS ISOSCELES.
TAKEAWAY #1: if they show a triangle containing equal angles, then it's the resultant equal sides that are important. if they show you a triangle containing equal sides, then it's the resultant equal angles that are important.
first, you should be able to eliminate choices (b) and (d) fairly quickly, because it should be fairly clear that statement 2 is insufficient. this is the case because, if only statement 2 is true, then you have NO lengths to work with at all; the sides of the figure could be numbers in the millions, tiny decimals, or anything in between. this situation is clearly insufficient for determining the length of one of the sides.
TAKEAWAY #2: you can't determine lengths/sizes unless you know the length/size of at least one thing in the picture.
now examine statement one.
here are two different approaches:
1. ALGEBRAIC APPROACH: (this is the preferred approach - this algebra is straightforward, albeit a bit tricky, but none of it requires any creative thinking at all)
just take advantage of the fact that there are 180 degrees in a triangle, and also 180 degrees in a straight line.
TAKEAWAY #3: almost every problem mixing variables and angles comes down to exactly the same two things: 180° in a triangle, and 180° in a straight line.
look at angle ADB. since the angle adjacent to it is 2x°, and the two angles together make a straight line, angle ADB must be (180 - 2x)°.
now look at angle ABD. this is a bit trickier, but you use the fact that a triangle has 180 degrees. therefore, add the other two angles together, and then subtract that total from 180°.
ABD = 180° - (x° + (180 - 2x)°)
= 180° - (180° - x°)
= 180° - 180° + x°
= x°
TAKEAWAY #4: don't forget to distribute subtraction when you have quantities in parentheses.
therefore, triangle ABD is isosceles.
therefore, AD = BD. but we already know BD = BC, so AD = BD = BC = 6.
sufficient.
approach #2 follows.[/list]
Ron has been teaching various standardized tests for 20 years.
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