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ok can't figure out this one.
The numbers given in the problem and the diagram you posted don't seem to go with the answer choices. I'm also not sure how to solve this one, but something seems mixed up also...
that's what the choices are for this question /:
yeah, I'm not doubting you... I just don't know what to do with the situation.
AC/AD = AE/AB
So, AD is 5
how did you know to set up a ratio of the short portion of one leg to the whole length of the other leg?
There is a good website explaining all this. Look at example 2: http://www.pinkmonkey.com/studyguides/subjects/geometry/chap7/g0707701.asp
It goes with figure 7.21b which is the figure on the right.
They use different letters for their points, so you have to rewrite the letters, but its almost the exact same problem.
It's corresponding sides of similar triangles
Very nice... I was thinking ratios for similar triangles, but I didn't catch the fact that the "slanted" base and midsections would cause the ratios to be different. Thanks for the link... my goal of "learn something new daily" has been met :)
This is sort of obscure, but geometry is way under-rated. Excellent way to build logic.
A little bit more about the cyclic quadrilateral:
In that website that I referenced, they alluded to properties of the cyclic quadrilateral and that might have left some here a little empty like it was magic "balck box" stuff. One really has to come to grips with the cyclic quadrilateral first, so here's another website that explains why I could use that ratio: http://www.onlinemathlearning.com/quadrilateral-circle.html
It's a video (near the bottom) called "Proof for the Cyclic Quadrilateral"