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Let \(ABC\) be a triangle with \(\angle BAC=120^{\circ}\). Let \(D\) be the midpoint of \(BC\) and suppose that \(AD\) is perpendicular to \(AB\). Let \(E\) be the second point of intersection of the line \(AD\) with the circumcircle of triangle \(ABC\). Let \(F\) be the intersection of line \(AB\) and \(CE\).

(a) Prove that line \(DF\) is perpendicular to line \(BE\).

(b) Prove that \(DF = BC\).

Note by Sharky Kesa
1 year, 10 months ago

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Yes, It is very obvious.

\([1]\) In \(\Delta BEF\), \(BC \perp EF\) and \(EA \perp BF.\) => \(D\) is the orthocenter. => \(DF \perp BE\).

\([2]\) Clearly, \(\angle CEB = 60^{\circ} = \angle CDF \) => In right \(\Delta FCD\) , \(\dfrac{DF}{DC} = sin 30^{\circ} = \dfrac{2}{1}\) => \(\dfrac{DF}{BC} = \dfrac{2}{2} = \dfrac{1}{1}\) Rohit Camfar · 5 months, 1 week ago

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@Sharky Kesa : I am able to prove only the part (a). My proof is : It can be observed that D is the orthocentre of triangle BFE. Hence the result follows. Shrihari B · 1 year, 9 months ago

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I am solving thia Aditya Singh · 1 year, 9 months ago

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@Aditya Singh Still solving? Sharky Kesa · 1 year, 9 months ago

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