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Congruent triangles will be difficult to prove in this case. Consider the case that angle DAB is right while angle ABC is obtuse. The only condition on this quadrilateral is that two sides are parallel - don't take more from the picture than is stated. Using the formula for area might be a better way of going about it. --[[User:Trjacobs|Trjacobs]] 20:31, 2 September 2009 (UTC) | Congruent triangles will be difficult to prove in this case. Consider the case that angle DAB is right while angle ABC is obtuse. The only condition on this quadrilateral is that two sides are parallel - don't take more from the picture than is stated. Using the formula for area might be a better way of going about it. --[[User:Trjacobs|Trjacobs]] 20:31, 2 September 2009 (UTC) | ||
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| + | Is it possible to draw in the heights for each triangle and then use side DC and a common base, show that the hieghts are equal, and then just state that the areas are equal because of the (1/2)base*height? | ||
Revision as of 03:17, 3 September 2009
Can anyone figure out how to do this problem in an effective way that doesn't take 30 steps? :)
I used the new BF 16 - Congruent figures have the same area. Still to be seen if this is correct.
I want to know how you even got to conguent triangles. I feel like I need more information to even get that far. --Shore 19:15, 2 September 2009 (UTC)
Congruent triangles will be difficult to prove in this case. Consider the case that angle DAB is right while angle ABC is obtuse. The only condition on this quadrilateral is that two sides are parallel - don't take more from the picture than is stated. Using the formula for area might be a better way of going about it. --Trjacobs 20:31, 2 September 2009 (UTC)
Is it possible to draw in the heights for each triangle and then use side DC and a common base, show that the hieghts are equal, and then just state that the areas are equal because of the (1/2)base*height?
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