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Question figure

In $\Delta ABC$, $\angle EAF=10^{\circ}$, $\angle FAB=70^{\circ}$, $\angle FBA=60^{\circ}$, $\angle DBF=20^{\circ}$. The goal is to find $\angle EDA$.

I have tried this question for some time and am unable to find the required angle. On manually measuring angles, I found that $\angle EDA = 20^{\circ}$, which implies that $\Delta CDE\sim \Delta BFD$. I have tried to prove this result using sine rule but got nowhere. Also, I have checked and there are no cyclic quadrilaterals in this configuration. So, there might be a clever construction that solves this problem. Any solution will be much appreciated. Thanks a lot. Please try not to use values of $\cos 20$ and such in your solution as this may not be provided in contest math.

Pravimish
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  • Take a look at the recent solution I have given here by embedding this triangle into a stellated polygon. – Jean Marie Jun 11 '21 at 11:32
  • @Jean Marie, the question has differing constraints as compared to mine. The question in the link provides the condition, $CD=AE=AB$(According to my figure), but this is not necessary according to the constraints in my question(given only angles). Please clarify – Pravimish Jun 11 '21 at 12:30
  • @MatthewDaly, this question requires the knowledge of cosines of angles like $20^{\circ}$ and this is contest math, where the values may not be provided – Pravimish Jun 11 '21 at 12:33
  • Here is a page dedicated to this problem and all its variants: https://www.cut-the-knot.org/triangle/80-80-20/index.shtml – Intelligenti pauca Jun 11 '21 at 13:27

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