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Math
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2 \times 3
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234
a_{i-1}
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\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
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\boxed{123}
123
Comments
no it isnt . We can say that FCD is an equilateral triangle as well.(Since AB is parallel to FC and i forgot which theorem it is, but keep my word for it). Now we can easily figure out that angle BFC is 30 degrees . Since total angle DFB is 90 degrees, we can say that it is not angle bisector
Sorry that I replied this late. Not until now have I realized this problem is very silly. Indeed, AB is parallel to FC. In a equilateral triangle, the bisector and the median of one angle is the same one. AC is the angular bisector of ∠BAD. Hence, CD is half of BD. BF is the median of AD, so FD half of AD. Since BD=AD, therefore CD=FD. Together with "∠D=60°", we can get FCD is an equilateral triangle. Thanks for your patience of answering such a simple question. I will think more before asking in the future.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
no it isnt . We can say that FCD is an equilateral triangle as well.(Since AB is parallel to FC and i forgot which theorem it is, but keep my word for it). Now we can easily figure out that angle BFC is 30 degrees . Since total angle DFB is 90 degrees, we can say that it is not angle bisector
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Sorry that I replied this late. Not until now have I realized this problem is very silly. Indeed, AB is parallel to FC. In a equilateral triangle, the bisector and the median of one angle is the same one. AC is the angular bisector of ∠BAD. Hence, CD is half of BD. BF is the median of AD, so FD half of AD. Since BD=AD, therefore CD=FD. Together with "∠D=60°", we can get FCD is an equilateral triangle. Thanks for your patience of answering such a simple question. I will think more before asking in the future.