Prove AB=2CE | Geometry Problem

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To prove that AB=2CE, start by calculating the angles of triangle CDE in terms of angle BAC, given that line AC is parallel to DE. Use the sine formula to establish a relationship between the lengths |CE| and |CD|. The varying angle BAC will affect the angles of triangle CDE, but they remain interrelated. This relationship allows for the necessary proof to be constructed. The discussion emphasizes the importance of understanding angle relationships in geometry.
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In the figure attached, I have to prove AB=2CE given that: line AC is parallel to DE and angles as mentioned in the figure. Can anyone please help me to prove the same?
 

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Calculate angles of the triangle ##CDE## in terms of ##\angle BAC## Then use the sine formula to relate ##|CE|## to ##|CD|##
 
∠BAC can vary and so will the angles of the Triangle CDE
 
rushikesh said:
∠BAC can vary and so will the angles of the Triangle CDE

They can. But they're related, so you can express one in terms of the other.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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